Question

To verify her suspicion that a rock specimen is hollow, a geologist weighs the specimen in air and in water. She finds that the specimenweighs twice as much in air as it does in water. The density of the solid part of the specimen is 6.32 x 103 kg/m³. What fraction of thespecimen's apparent volume is solid?NumberUnits

Answer 1

Given:

Density of the solid part = 6.23 x 10³ kg/m³

Let's find the fraction of the specimen's apparent volume that is solid.

Apply the formula:

$\frac{V_s}{V_r}=\frac{2\rho_w-\rho_a}{\rho_s-\rho_a}$

Where:

Vs is the volume of solid in rock

Vr is the volume of rock.

ρa = 1.275 kg/m³

ρw is the density of water = 1000 kg/m³

ρs is the density of solid rock = 6.23 x 10³ kg/m³

Plug in the values in the equation and solve.

We have:

$\begin{gathered} \frac{V_s}{V_r}=\frac{2*1000-1.275}{6.23*10^3-1.275} \\ \\ \frac{V_s}{V_r}=\frac{2000-1.275}{6230-1.275}=\frac{1998.725}{6228.725} \\ \\ \frac{V_s}{V_r}=0.32 \end{gathered}$

Therefore, the fraction of the specimen's apparent volume that is solid is 0.32

ANSWER:

0.32