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.

Answer 1

Given:
Specific gravity of water, ${\rho }_W$ = 1 gm/cc
Specific gravity of ice, ρ_ice = 0.9 gm/cc
Specific gravity of kerosene oil, ρ_k = 0.8 gm/cc
Now,
V_​ice = V_k + V_w
Here,
V_k = Volume of ice inside kerosene oil
V_w = Volume of ice inside water
V_ice = Volume of ice
Thus, we have:
$$\begin{gathered} V_{\mathrm{ice}}\times {\mathrm{\rho }}_{\mathrm{ice}}\times g=V_{\mathrm{k}}\times {\rho }_{\mathrm{k}}\times g+V_{\mathrm{w}}\times {\rho }_{\mathrm{w}}\times g \\ \Rightarrow \left(V_k+V_w\right)\times {\rho }_{\mathrm{ice}}=V_{\mathrm{k}}\times {\rho }_{\mathrm{k}}+V_{\mathrm{w}}\times {\rho }_{\mathrm{w}} \\ \Rightarrow (0.9)V_{\mathrm{k}}+(0.9)V_{\mathrm{w}}=(0.8)V_{\mathrm{k}}+\left(1\right)\times V_{\mathrm{w}} \\ \Rightarrow (0.1)V_{\mathrm{w}}=0.1V_{\mathrm{k}} \\ \Rightarrow \frac{V_{\mathrm{w}}}{V_{\mathrm{k}}}=1. \\ \Rightarrow V_{\mathrm{w}}:V_{\mathrm{k}}=1:1 \end{gathered}$$