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 $$= 0.9$$)

Let $$V_w$$ be the volume of ice in water and $$V_k$$ be the volume in kerosene. 

The total volume of the ice is $$V = V_w + V_k$$.

$$\text{Weight of Ice} = \text{Buoyant force (Water)} + \text{Buoyant force (Kerosene)}$$

$$\rho_i (V_w + V_k)g = \rho_w V_w g + \rho_k V_k g$$

Using specific gravities ($$\rho_w = 1$$, $$\rho_i = 0.9$$, and $$\rho_k = 0.8$$):

$$0.9(V_w + V_k) = 1(V_w) + 0.8(V_k)$$

$$0.9V_w + 0.9V_k = 1V_w + 0.8V_k$$

$$0.1V_k = 0.1V_w$$

$$\frac{V_w}{V_k} = \frac{0.1}{0.1} = \frac{1}{1}$$

The ratio of the volume immersed in water to that in kerosene is $$1 : 1$$.