# 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:

$\ \\ \rho_1 = 7850 kg/m^3 \ \\ \rho_2 = 1000 kg/m^3 \ \\ \ \\ \dfrac12 \rho_2 \dfrac43 \pi r^3 = 0.5 \ kg \ \\ r = \sqrt{\dfrac34 \cdot 2\cdot 0.5 / \pi / 1000 } \ \\ r = 0.062035 \ m = 6.2 \ \text{ cm } \ \\ \ \\ \rho_1 \dfrac43 \pi r^3 - \rho_1 \dfrac43 \pi (r-h)^3 = 0.5 \ kg \ \\ \ \\ r^3-(r-h)^3 = \dfrac{0.5 \cdot 3}{ \rho_1 4 \pi } \ \\ (r-h)^3 = r^3 - \dfrac{0.5 \cdot 3}{ \rho_1 4 \pi } \ \\ r -h = \sqrt{ r^3 - \dfrac{0.5 \cdot 3}{ \rho_1 4 \pi } } \ \\ h = r - \sqrt{ r^3 - \dfrac{0.5 \cdot 3}{ \rho_1 4 \pi } } \ \\$
$h = 0.135 \ \text{ cm }$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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