A cylindrical cup is 8 cm in diameter and 20 cm tall and is filled halfway with saltwater. The density of the saltwater is 1.1 g/cm^3.
A spherical ball has a diameter of 4 cm and a mass of 40g. How much does the level of the water rise when the ball is dropped into the cup?

Volume of sphere is 4/3 pi r cubed, where r is 2cm, equals 33.51 cubic cms.  Density of sphere is mass/volume = 1.19366 g/cm3.  Shere is more dense than the salt water and therefore the sphere sinks (as per the title of the question).

The sphere will therefore displace it's volume of saltwater (not it's mass)

Surface area of flask is pi r squared, where r = 4cm, equals 50.2655 square centimetres.  Volume of displaced water will be surface area times rise in water level, therefore rise in water level equals volume of sphere divided by surface area of flask, equals 0.66666cm (2/3 of a centimetre).

Posted by Fletch on 2006-01-03 09:32:06