Question

Bowling balls are roughly the same size but come in a variety of weights. Given its official circumference of roughly 27.0 in calculate the heaviest bowling ball that will float in a fluid of specific gravity 1.200.

Answer 1

Specific gravity of 1.2 means the density is 1.2 times that of water, or

density of fresh water at 20C = 0.998 g/cm³

     = 998 kg/m³ = 8.33 lb/gal = 62.1 lb/ft³

in this case, density = 1.2 x 62.1 lb/ft³ = 74.52 lb/ft³

circumference of 27 inch = diameter of 27/π and radius of 13.5/π

Volume = ⁴/₃πr³ = ⁴/₃π(13.5/π)³ = 4•13.5/π²

converting to ft³, that is 4•(13.5)³/(3π²12³) = 0.192 ft³

to just float the density is as above

density = mass/volume

mass = density x volume = 74.52 lb/ft³ *0.192 ft³

= 14.3 lb