A cube of water ice ($\rho$= 0.917 * 103 kg/m3) is placed in mercury ($\rho$= 13.6 *103 kg/m3), which is liquid at 0 $\degree$ Celsius. If we ignore any possible melting of the ice cube and problems with the surface tension of mercury, the fraction of the ice cube that floats above the surface of the mercury is:
A)0.067 4.
B)0.073 5.
C)0.926.
D)0.933.
E)1.00.

Question

Quizplus · Accepted Answer

The fraction of the ice cube that floats above the surface is determined by the ratio of the densities: $1 - \frac{\rho_{ice}}{\rho_{mercury}} = 1 - \frac{0.917}{13.6} \approx 0.933$.

A Cube of Water Ice ρ\rhoρ = 0917 * 103 Kg/m3) Is Placed in Mercury

A Cube of Water Ice $\rho$ = 0917 * 103 Kg/m3) Is Placed in Mercury