Asked by David
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?
Answers
Answered by
drwls
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.
(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.
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