Consider a solid sphere of radius R and mass density  $\rho \left(r\right)={\rho }_0\left(1-\frac{r^2}{R^2}\right)$,  $0<r\le R$. The minimum density of a liquid in which it will float is:

$$\begin{gathered} Given : \rho ={\rho }_0\left(1-\frac{r^2}{R^2}\right) for 0<r\le R \\ Consider a an element of thickness dr, at a dis\tan ce r from center. \\ Its volume is dV = \left(4\pi r^2\right)dr \\ Its mass is dm=\rho dV = {\rho }_0\left(1-\frac{r^2}{R^2}\right) \left(4\pi r^2\right)dr \\ Integrating, \\ m=\int dm=\underset{0}{\overset{R}{\int }}{\rho }_{0}4\pi \left(r^2-\frac{r^4}{R^2}\right)dr={\rho }_{0}4\pi {\left({\frac{r}{3}}^3-\frac{r^5}{5R^2}\right)}_0^R=\frac{8}{15}{\rho }_0\pi R^3 \\ To float in liquid , mg=B \\ \Rightarrow \frac{8}{15}{\rho }_0\pi R^{3}g={\rho }_L\frac{4}{3}\pi R^{3}g \\ \Rightarrow \frac{2}{5}{\rho }_0={\rho }_L \end{gathered}$$