As shown in the figure, a cube of ice floats partly in water and...

Let the total volume of the ice cube be $V$.

Let $V_w$ and $V_k$ be the volumes of the ice cube immersed in water and K.oil respectively.

Since the ice cube is completely submerged in the two liquids, $V = V_w + V_k$.

For the ice cube to float in equilibrium, its total weight must be balanced by the total buoyant force exerted by the water and K.oil.

Total weight of ice = Buoyant force due to water + Buoyant force due to K.oil

$V \rho_i g = V_w \rho_w g + V_k \rho_k g$

Substituting the specific gravities (which are proportional to densities):

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

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

$0.9 V_k - 0.8 V_k = V_w - 0.9 V_w$

$0.1 V_k = 0.1 V_w$

$\dfrac{V_w}{V_k} = \dfrac{1}{1}$

The ratio of the volume of ice immersed in water to that immersed in K.oil is $1 : 1$.