# Hollow sphere

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 density of steel is 7850 kg/m3

Result

r =  6.2 cm
h =  0.135 cm

#### Solution:

$\ \\ \rho_1 = 7850 kg/m^3 \ \\ \rho_2 = 1000 kg/m^3 \ \\ \ \\ \dfrac12 \rho_2 \dfrac43 \pi r^3 = 0.5 \ kg \ \\ r = \sqrt[3]{\dfrac34 \cdot 2\cdot 0.5 / \pi / 1000 } \ \\ r = 0.062035 \ m = 6.2 \ \text{ cm } \ \\ \ \\ \rho_1 \dfrac43 \pi r^3 - \rho_1 \dfrac43 \pi (r-h)^3 = 0.5 \ kg \ \\ \ \\ r^3-(r-h)^3 = \dfrac{0.5 \cdot 3}{ \rho_1 4 \pi } \ \\ (r-h)^3 = r^3 - \dfrac{0.5 \cdot 3}{ \rho_1 4 \pi } \ \\ r -h = \sqrt[3]{ r^3 - \dfrac{0.5 \cdot 3}{ \rho_1 4 \pi } } \ \\ h = r - \sqrt[3]{ r^3 - \dfrac{0.5 \cdot 3}{ \rho_1 4 \pi } } \ \\$
$h = 0.135 \ \text{ cm }$

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