Sphere With Cavity

Assuming the mass of air is negligible and the cavity is a vacuum, then the mass of the metallic body is $m_m = 40 \text{ g}$ . Since the density of the metal is $\rho = 8 \text{ g/cm}^3$ , then the volume of the metal body is $\dfrac {m_m}\rho = 5 \text{ cm}^3$ . When the sphere is weighed in water, the weight reduced $20 \text{ g}$ is the weight of water displaced and $20 \text{ g}$ of water is $20 \text{ cm}^3$ . Therefore, the volume of the sphere is $20 \text{ cm}^3$ , minus the $5 \text{ cm}^5$ metal shell, we have the volume of the cavity of $\boxed{15} \text{ cm}^3$ .