(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?

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

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

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

The apparent weight of a substance is,

\begin{aligned}{m_{{\rm{app}}}}g={m_{{\rm{ac}}}}g-{F_{\rm{B}}}\\{m_{{\rm{app}}}}g={m_{{\rm{ac}}}}g-V{\rho_{\rm{f}}}g\\{m_{{\rm{ac}}}}-{m_{{\rm{app}}}}=V{\rho _{\rm{f}}}\end{aligned}

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{aligned}{m_{{\rm{ac}}}}-{m_{{\rm{app}}}}=\frac{{{m_{{\rm{ac}}}}}}{{{\rho _{\rm{s}}}}}{\rho_{\rm{f}}}\\{\rho_{\rm{s}}}=\frac{{{m_{{\rm{ac}}}}}}{{{m_{{\rm{ac}}}}-{m_{{\rm{app}}}}}}{\rho _{\rm{f}}}\end{aligned}

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

Substituting the given values we get,

\begin{aligned}{\rho_{\rm{s}}}=\frac{{9.28\;{\rm{kg}}}}{{9.28\;{\rm{kg}}-6.18\;{\rm{kg}}}}\times1000\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\\\approx2.99\times{10^3}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\end{aligned}

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