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:

 ρ1=7850kg/m3 ρ2=1000kg/m3  12ρ243πr3=0.5 kg r=3420.5/π/10003 r=0.062035 m=6.2 cm  ρ143πr3ρ143π(rh)3=0.5 kg  r3(rh)3=0.53ρ14π (rh)3=r30.53ρ14π rh=r30.53ρ14π3 h=rr30.53ρ14π3  \ \\ \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 cmh = 0.135 \ \text{cm}



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