Hollow sphere

The hollow steel sphere floats on the water, plunged into half its volume. Determine the outer radius of the sphere and wall thickness if you know that the weight of the sphere is 0.5 kg and the density of steel is 7850 kg/m³

Correct answer:

r =  6.2 cm
h =  0.1346 cm

Step-by-step explanation:

$$\begin{gathered} {\rho }_1=7850kg/m^3 \\ {\rho }_2=1000kg/m^3 \\ \frac{1}{2}{\rho }_2\frac{4}{3}\pi r^3=0.5kg \\ r=\sqrt[3]{\frac{3}{4}\cdot 2\cdot 0.5/\pi /1000} \\ r=0.062035m=6.2\text{cm} \\ {\rho }_1\frac{4}{3}\pi r^3-{\rho }_1\frac{4}{3}\pi (r-h{)}^3=0.5kg \\ {r}^3-(r-h{)}^3=\frac{0.5\cdot 3}{{\rho }_{1}4\pi } \\ (r-h{)}^3=r^3-\frac{0.5\cdot 3}{{\rho }_{1}4\pi } \\ r-h=\sqrt[3]{r^3-\frac{0.5\cdot 3}{{\rho }_{1}4\pi }} \\ h=r-\sqrt[3]{r^3-\frac{0.5\cdot 3}{{\rho }_{1}4\pi }} \end{gathered}$$

$h=0.1346\text{cm}$

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