# Water has a density of 1 g/cm^3. You drop an object into the water that has a volume of 20 cm^3 and a mass of .86 g. What will happen to the object?

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### 1 Answer

When an object falls in water, it experiences buoyancy force acting on it from water. This force equal `F=rho_w*V_s*g` , where `rho_w` is the density of water, `V_s` is the volume submerged, and g is the gravitational acceleration. If the magnitude of the buoyancy force is smaller than the gravity, the object will drown. If the magnitude of the buoyancy force is greater than the gravity, the object will rise to the surface of the water, so that it will not be totally submerged. The submerged volume will therefore be less than the whole volume, the buoyancy force will become equal to gravity, and the object will float.

Therefore, to find out what happens to the object, we need to find its density. If the object's density is greater than that of water, the object will drown. Otherwise, it will float.

The density of the object can be found by dividing its mass by its volume:

`rho = m/V = (.86 g)/(20 cm^3) = 0.043 g/(cm^3)` .

This is smaller than the density of water, 1 g/cm^3, so **the object will float**.

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