Consider a solid sphere of radius R and mass density . The minimum density of a liquid in which it will float is:

Engineering

Physics

Liquid Pressure

Calculus

Question

Consider a solid sphere of radius R and mass density $ρ (r) = ρ_{0} (1 - \frac{r^{2}}{R^{2}}) 0 < r \leq R$ . The minimum density of a liquid in which it will float is:

$\frac{2 ρ_{0}}{3}$

$\frac{ρ_{0}}{3}$

$\frac{2 ρ_{0}}{5}$

$\frac{ρ_{0}}{5}$

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$ρ = ρ_{0} (1 - \frac{r^{2}}{R^{2}}) 0 < r \leq R$

mg = B

$\int ρ (4 {πr}^{2} dr) = ρ_{L} \frac{4}{3} {πR}^{3}; \int_{0}^{R} ρ_{0} (1 - \frac{r^{2}}{R^{2}}) 4 {πr}^{2} dr = ρ_{L} \frac{4}{3} {πR}^{3}$

$\int_{0}^{R} ρ_{0} 4 π (r^{2} - \frac{r^{2}}{R^{2}}) dr = ρ_{0} 4 π (\frac{r^{3}}{3} - \frac{r^{5}}{5 R^{2}}) = ρ_{L} \frac{4}{3} {πR}^{3}; \frac{2}{5} ρ_{0} = ρ_{L}$