Hollow sphere

The steel hollow 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 result:

r =  6.2 cm
h =  0.1346 cm

Solution:

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

$h=0.1346\text{ cm}$

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