The density of ice is 917 kg m^-3. What fraction of the volume of a piece of ice will be above when floating in freshwater?

Density of ice $\mathrm{\rho }=917 \mathrm{kg}{ \mathrm{m}}^{-3}$

$\text{ Density of fresh water, }\rho =1000 \mathrm{kg}{ \mathrm{m}}^{-3}$

Let V be the total volume of the ice and v be the volume of ice above the water. 

The volume of the water displaced by the immersed part of ice =(V-$\nu$)

Weight of ice = Weight of the water displaced 

$\mathrm{V}\times 917\times \mathrm{g}=(\mathrm{V}-\mathrm{v})\times 1000\times \mathrm{g}$

$\frac{\mathrm{v}}{\mathrm{V}}=\frac{83}{1000}=0.083$