Question

(II) A geologist finds that a Moon rock whose mass is 9.28kg has an apparent mass of 6.18kg when submerged in water. What is the density of the rock?

Short answer

The density of rock is 2.99 x 10³ kg/m³

Step-by-step solution

Understanding the apparent weight

The apparent weight inside a fluid is the difference between the actual weight and the buoyancy given by the fluid.

Given the data

The actual mass of the Moon rock is m_ac = 9.28kg.

The apparent mass of the rock ism_app = 6. 18kg.

Calculation the density of rock 

The apparent weight of a substance is,

$$\begin{array}{r}{m}_{\mathrm{a}\mathrm{p}\mathrm{p}}g=m_{\mathrm{a}\mathrm{c}}g-F_{\mathrm{B}}\\ m_{\mathrm{a}\mathrm{p}\mathrm{p}}g=m_{\mathrm{a}\mathrm{c}}g-V{\rho }_{\mathrm{f}}g\\ m_{\mathrm{a}\mathrm{c}}-m_{\mathrm{a}\mathrm{p}\mathrm{p}}=V{\rho }_{\mathrm{f}}\end{array}$$

Here, F_B is the buoyant force by the fluid, V is the submerged volume of the solid,ρ_f is the density of the fluid, and g is the acceleration due to gravity.

The density of rock isρ_s . Therefore,

$$\begin{array}{r}{m}_{\mathrm{a}\mathrm{c}}-m_{\mathrm{a}\mathrm{p}\mathrm{p}}=\frac{m_{\mathrm{a}\mathrm{c}}}{{\rho }_{\mathrm{s}}}{\rho }_{\mathrm{f}}\\ {\rho }_{\mathrm{s}}=\frac{m_{\mathrm{a}\mathrm{c}}}{m_{\mathrm{a}\mathrm{c}}-m_{\mathrm{a}\mathrm{p}\mathrm{p}}}{\rho }_{\mathrm{f}}\end{array}$$

Here,ρ_f = 1000kg/m³ is the density of water.

Substituting the given values we get,

$$\begin{array}{r}{\rho }_{\mathrm{s}}=\frac{9.28\mathrm{k}\mathrm{g}}{9.28\mathrm{k}\mathrm{g}-6.18\mathrm{k}\mathrm{g}}\times 1000\mathrm{k}\mathrm{g}/{\mathrm{m}}^3\\ \approx 2.99\times {10}^3\mathrm{k}\mathrm{g}/{\mathrm{m}}^3\end{array}$$

Hence, the density of the rock is 2.99 x 10³ kg/m³.