Question

Suppose that a solid cube of some material floats on the surface of an oil puddle at a service station with 60.3 % of its volume submerged. The density of the oil is 907 kg/m3.What is the density of the cube's material?

Answer 1

To solve this problem we will apply the principle of Archimedes which reflects the proportion between the submerged volume of one object in another and the density of their bodies. This relationship is given by the following function:

$\frac{\text{Sumerged volume of Solid}}{\text{Total Volume of Solid}} = \frac{\text{Density of solid}}{\text{Density of liquid}}$

Replacing with our values we have that,

$\frac{60.3V}{100V} = \frac{\rho}{907kg/m^3}$

$\rho = 907kg/m^3 * \frac{60.3}{100}$

$\rho = 546.921kg/m^3$

Therefore the density of the cube's material is $546.921kg/m^3$

Answer 2

The density of the cube's material is 546.921 Kg/m³

From the question given above, the following data were obtained:

Submerged volume of material = 60.3 m³

Total volume of material = 100 m³

Density of oil = 907 Kg/m³

Density of cube material =.?

The density of the cube material can be obtained as illustrated below:

$\frac{submerged \: volume}{total \: volume} = \frac{density \: of \: material}{density \: of \: liquid} \\ \\ \frac{60.3}{100} = \frac{density \: of \: material}{907} \\ \\ 0.603 = \frac{density \: of \: material}{907} \\ \\ cross \: multiply \\ \\ density \: of \: material = 0.603 \times 907$

Density of cube material = 546.921 Kg/m³

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