A cube of ice floats partly in water and partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil (specific gravity of Kerosene oil = 0.8, specific gravity of ice = 0.9): [figure]
- 5 : 4
- 9 : 10
- 8 : 9
- 1 : 1
Answer
(4) 1 : 1
Solution
Let Vw, Vk = volumes of the cube in water and in kerosene (cube fully inside the liquids).
Since the ice floats, its weight = upthrust of water + upthrust of kerosene:
(Vw + Vk)(0.9) g = Vw(1) g + Vk(0.8) g
0.9Vw + 0.9Vk = Vw + 0.8Vk
0.1Vk = 0.1Vw
Vw : Vk = 1 : 1
Answer: (4)