Consider a solid sphere of radius R and mass | vedclass

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(D) For the sphere to float with minimum liquid density,it must be completely submerged. In this state,the weight of the sphere equals the buoyant for

Vedclass · Accepted Answer

$\frac{2\rho_{0}}{5}$

Vedclass · Answer

(D) For the sphere to float with minimum liquid density,it must be completely submerged. In this state,the weight of the sphere equals the buoyant force.Weight $W = mg = \int ho(r) g dV = \int_{0}^{R} ho_{0} \left(1 - \frac{r^{2}}{R^{2}}ight) g (4 \pi r^{2} dr)$.$W = 4 \pi ho_{0} g \int_{0}^{R} \left(r^{2} - \frac{r^{4}}{R^{2}}ight) dr$.$W = 4 \pi ho_{0} g \left[ \frac{r^{3}}{3} - \frac{r^{5}}{5R^{2}} ight]_{0}^{R} = 4 \pi ho_{0} g \left( \frac{R^{3}}{3} - \frac{R^{3}}{5} ight) = 4 \pi ho_{0} g \left( \frac{2R^{3}}{15} ight) = \frac{8}{15} \pi R^{3} ho_{0} g$.Buoyant force $B = V_{sphere} ho_{l} g = \frac{4}{3} \pi R^{3} ho_{l} g$.Equating $W = B$: $\frac{8}{15} \pi R^{3} ho_{0} g = \frac{4}{3} \pi R^{3} ho_{l} g$.Solving for $ho_{l}$: $ho_{l} = \frac{8}{15} 	imes \frac{3}{4} ho_{0} = \frac{2}{5} ho_{0}$.

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

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