Question

3. The density of bubble-filled ice from your refridgerator is 900 kg/m?. You have an ice cube that is 28 mm wide x 40 mm long x 10 mm high which floats in a large glass of a sugar-sweetened drink with a density of 1.05 g/cm". Part of the ice pokes up above the liquid. a) What is the cube's weight F? b) What volume Vdisp of drink displaced by the ice would be needed to provide a buoyant force the same as the cube's weight? c) What fraction of the ice cube is below the surface (i.e. the ratio of displaced liquid to ice volume)? Your answer should be a rational number with single digits both in the numerator and in the denominator... Note: 1 m - 100 cm - 1000 mm, but for areas 1 m² = (1 m)2 = (1000 mm)? # 1000 mm and for volumes 1 m² = (1 m) = (1000 mm) 1000 mm!]

Answer 1

Density of ice $\rho_i=900\,kg/m^3$

Density of drink $\rho_f=1050\,kg/m^3$

Volume of cube

V = 28 * 40 * 10 mm = 11.2 * 10-6 m3

a)

Cube's weight $F_g=mg=(\rho _i V)g=900*11.2*10^{-6}*9.81=0.099\,N$

b)

Buoyant force

FB = F,

$\rho_f V_{dis}g=mg$

$V_{dis}=\frac{m}{\rho_f}=\frac{\rho _i V}{\rho_f}=\frac{900*11.2*10^{-6}}{1050}=9.6*10^{-6}\,m^3$

c)

$\frac{V_{dis}}{V}=\frac{9.6*10^{-6}}{11.2*10^{-6}}=0.857$

Fraction of ice cube below the surface of drink is