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

Any point inside the sphere has density equal to

$\mathrm{p}\left(\mathrm{r}\right)={\mathrm{p}}_0\left[1-\frac{{\mathrm{r}}^2}{{\mathrm{R}}^2}\right]$

where r denotes the point's distance from the centre.

As a result, the density will be the same for all spots that are equally spaced from the centre.

A similar elemental shell with thickness dr weighs around

$\mathrm{dM}=\mathrm{p}\left(\mathrm{r}\right)\mathrm{dV}=\mathrm{p}\left(\mathrm{r}\right)4{\mathrm{\pi r}}^2\mathrm{dr}$

$={\mathrm{p}}_0\left(1-\frac{{\mathrm{r}}^2}{{\mathrm{R}}^2}\right)4{\mathrm{\pi r}}^2\mathrm{dr}$

$=4{\mathrm{\pi p}}_0\left(1-\frac{{\mathrm{r}}^2}{{\mathrm{R}}^2}\right){\mathrm{r}}^2\mathrm{dr}$

$\mathrm{Total} \mathrm{mass} \mathrm{M}=\int \mathrm{dM}=4\mathrm{\pi }{\int }_0^{\mathrm{k}}\left[{\mathrm{r}}^2-\frac{{\mathrm{r}}^4}{{\mathrm{R}}^2}\right]\mathrm{dr}$

$=4{\mathrm{\pi p}}_0{{\left[\frac{{\mathrm{r}}^3}{3}-\frac{{\mathrm{r}}^5}{5{\mathrm{R}}^2}\right]}^{\mathrm{R}}}_0=\frac{8{\mathrm{\pi pR}}^3}{15}$

The sphere is simply submerged and suspended when the liquid reaches the required density.

Weight = buoyant force.

$\mathrm{Mg}=\mathrm{\sigma Vg},$

where V is the sphere's volume and is the liquid's density.

$\frac{8{\mathrm{\pi p}}_0{\mathrm{R}}^3}{15}\mathrm{g}=\mathrm{\sigma }\frac{4}{3}{\mathrm{\pi R}}^3\mathrm{g}$

$\Rightarrow \frac{8{\mathrm{p}}_0}{15}=\frac{\mathrm{\sigma }4}{3}\Rightarrow \mathrm{\sigma }=\frac{2{\mathrm{p}}_0}{5}$

Hence the correct answer is $\frac{2{\mathrm{p}}_0}{5}.$