Question

A cube of ice floats partly in water and partly in K.oil (figure 13-E5). 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.

Answer 1

 Let the volume of ice in water = X m³ and the volume of ice in K.oil = Y m³ 

So total weight of the ice cube =(X+Y)*900 kg 

{Because density of the ice = 0.9*1000 =900 kg/m³} 

In the floating condition, the total force of buoyancy by the water and the K.oil will be equal to the weight of the ice cube. 

The total force of buoyancy  = X*1000 + Y*800 kg 

Equating the two we get, 

(X+Y)*900 = X*1000+Y*800 

→1000X-900X = 900Y-800Y 

→100X = 100Y 

→X = Y 

So the volume of ice cube immersed in water is equal to the volume immersed in K.oil. Hence their ratio 

= 1:1.

Answer 2

The correct answer is-