An aluminum cylinder with a density of 2.70 g/cm3 has a volume of 3.8 ¡¿ 10-5 m3. If the cylinder is suspended from a scale and immersed in an unknown fluid, the scale reads 0.581 N? What is the specific gravity of the fluid?

Answer 1

The actual mass of the cylinder is
(density)*(volume)
= 2700 kg/m^3*3.8*10^-5 m^3
= 0.1026 kg
and its weight is M g = 1.007 N

The reduction in weight when immersed, 0.426 N, equals the buoyancy, which is the displaced fluid weight.

The density of the fluid is given by
(density)*g*V = 0.426 N
density = 1.143*10^3 kg/m^3

Divide that by the density of water for the specific gravity of the immersion fluid.