Question

A cube of ice floats partly in water and partly in K.oil. Find the ratio of the volume of ice immersed in water to that in K.oil. Specific gravity of K.oil is 0.8 and that of ice is 0.9.
Figure

Answer 1

The volume ratio is 1: 1.

Ice density  = $900 kg/m^3$

Explanation:

Step 1:

Let the ice volume in water be m3, and let the ice volume in K.oil be b m3

Therefore, Total Volume = a + b

Ice density  = Water density at 4°C Specific gravity.

Ice Density  = $0.9 \times 1000 kg/m^3$

Ice density  = $900 kg/m^3$

Step 2:

Therefore,  

total mass of the ice cube = $(a + b) \times 900 kg$

Theretofore, Its weight = $(a + b) \times 900 \times 10$

Step 3:

By the Flotation Rule, The combined water and kerosene oil buoyancy power shall equal the weight of the ice cube.

Total buoyancy force = $a \times 1000 + Y \times 800 kg$

$(a + b) \times 900 = a \times 1000 + b \times 800$

900a + 900b = 1000a + 800b

100a = 100b

a = b

Therefore, the volume ratio is 1: 1.