Question

A solid plastic cube with uniform density (side length 0.5 m) of mass 100 kg is placed in a vat of fluid whose specific gravity is 1.2. What fraction of the cube's volume floats above the surface of the fluid? O C. 2/3 O 1.45

Answer 1

Answer:

0.042 m^3

Explanation:

side, a = 0.5 m, m = 100 kg

density of fluid, d = 1.2 g/cm^3 = 1200 kg/m^3

Volume of cube, V = side^3 = (0.5)^3 = 0.125 m^3

density of solid = mass of solid / volume of solid cube

density of solid, D = 100 /  0.125 = 800 kg/m^3

Let v be the volume of cube immersed in fluid.

According to the principle of flotation

Weight of cube = Buoyant force acting on the cube

V x D x g = v x d x g

0.125 x 800 = v x 1200

v = 0.083 m^3

Volume above the level of fluid = V - v = 0.125 - 0.083 = 0.042 m^3