Question
Question: A cube of ice floats partially in water and partially in K.oil. Calculate the ratio of the volume of...
A cube of ice floats partially in water and partially in K.oil. Calculate the ratio of the volume of ice immersed in water to that in K.oil. The specific gravity of ice is 0.9 and that of K.oil is 0.8.
Solution
The density of the ice cube will be the product of the density of water at 4∘Cand the specific gravity. Using this find the total mass of the cube. The find the total weight of the ice cube. By the law of the Flotation, the total buoyant force by the water and the Kerosene oil will be equivalent to the weight of the ice cube. Substitute the values in it. This will help you in answering this question.
Complete step by step answer:
Let us assume that the volume of ice in water be am3 and the volume of ice in K.oil be bm3. Hence the total volume can be written as,
V=a+b
The density of the ice cube will be the product of the density of water at 4∘Cand the specific gravity. That is we can write that,
dice=dw×ρ
Where ρ be the specific gravity of the ice. This is given as,
ρ=0.9
And the specific gravity of oil is given as,
ρ′=0.8
And the density of water at 4∘Cwill be,
dw=1000kgm−3
Substituting the values in it,
dice=0.9×1000=900kgm−3
Hence, the total mass of the cube will be,
m=dice×V
That is,
m=(a+b)×900
The total weight of the cube will be,
W=(a+b)×900×10
By the law of the Flotation, the total buoyant force by the water and the Kerosene oil will be equivalent to the weight of the ice cube.
Therefore we can write that,
total buoyant force=a×1000+b×800
Total buoyant force can be written as,
total buoyant force=(a+b)×900
Substituting this in the equation will give,
(a+b)×900=a×1000+b×800
Rearranging the equation will give,