Answer to Question #124133 in Physics for ju

Question

A hollow sphere of inner radius r and outer radius R floats half submerged in a liquid with density (P). If the density of the material of which the sphere is made is Pi, find an expression of the inner radius r in terms of R, Pi and P

Answer 1

The volume submerged under water (water density is $\rho$) is half the total volume of the sphere:

$$V=\frac{V_R}{2}=\frac{2}{3}\pi R^3.$$

The buoyancy force for this condition is

$$F_b=\rho gV=\frac{2}{3}\pi R^3\rho g.$$

The force of gravity acting on the sphere consists of the force of gravity of the sphere's material (material density is $\rho_i$) assuming that there is vacuum inside:

$$F_g=mg=\rho_iV_mg,$$

where $V_m$ - volume of the sphere's material:

$$V_m=V_R-V_r=\frac{4}{3}\pi(R^3-r^3).$$

Thus:

$$F_g=\frac{4}{3}\pi(R^3-r^3)\rho_i g.$$

Since the sphere is floating, $F_b=F_g$:

$$\frac{2}{3}\pi R^3\rho g=\frac{4}{3}\pi(R^3-r^3)\rho_i g,\\\space\\ r=R\sqrt[3]{1-\frac{\rho}{2\rho_i}}.$$