Question

An aluminum cylinder weighs 1.03 N. When this same cylinder is completely submerged in alcohol, the volume of the displaced alcohol is 3.90 x 10-5 m3. If the cylinder is suspended from a scale while submerged in the alcohol, the scale reading is 0.730 N. What is the specific gravity of the alcohol?

Answer 1

In this case the displaced fluid weighs 1.03 - 0.730 = 0.300 N (the buoyancy force). The displaced fluid mass is 0.300/g = 0.0306 kg. Divide that by the cylinder's volume for the mass density
(demsity) = 0.0306/3.9*10^-5
= 784 kg/m^3

Divide that by ths density of water (1000 kg.m^3) to get the specific gravity of the alcohol.