A cube of water ice (ρ = 0.917 × 10³ kg/m³) is placed in mercury (ρ = 13.6 × 10³ kg/m³), which is liquid at 0° 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 of the mercury can be calculated using the principle of buoyancy, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. The fraction of the volume of the ice cube that is submerged is equal to the ratio of the density of ice (0.917 × 10^3 kg/m^3) to the density of mercury (13.6 × 10^3 kg/m^3). Thus, the fraction submerged is 0.917/13.6 = 0.0674, and the fraction that floats above is 1 - 0.0674 = 0.9326, which rounds to 0.933.

A Cube of Water Ice (ρ = 0