A cube of ice floats partly in water and partly in K.oil (in the following figure). 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.
Solution
Given:
Specific gravity of water, \[\rho_W\] = 1 gm/cc
Specific gravity of ice, ρice = 0.9 gm/cc
Specific gravity of kerosene oil, ρk = 0.8 gm/cc
Now,
Vice = Vk + Vw
Here,
Vk = Volume of ice inside kerosene oil
Vw = Volume of ice inside water
Vice = Volume of ice
Thus, we have:
\[V_{ice} \times \rho_{ice} \times g = V_k \times \rho_k \times g + V_w \times \rho_w \times g\]
\[ \Rightarrow \left( V_k + V_w \right) \times \rho_{ice} = V_k \times \rho_k + V_w \times \rho_w \]
\[ \Rightarrow (0 . 9) V_k + (0 . 9) V_w = (0 . 8) V_k + \left( 1 \right) \times V_w \]
\[ \Rightarrow (0 . 1) V_w = 0 . 1 V_k \]
\[ \Rightarrow \frac{V_w}{V_k} = 1 . \]
\[ \Rightarrow V_w : V_k = 1: 1\]
