Consider a solid sphere of radius $R$ and mass density $\rho(r)=\rho_{0}\left(1-\frac{r^{2}}{R^{2}}\right), 0 liquid in which it will float is:
Correct Option: , 3
(3) For minimum density of liquid, solid sphere has to float (completely immersed) in the liquid.
$m g=F_{B}\left(\right.$ also $\left.V_{\text {immersed }}=V_{\text {total }}\right)$
or $\int \rho d V=\frac{4}{3} \pi R^{3} \rho_{\ell}$
$\left[\rho(r)=\rho_{0}\left(1-\frac{r^{2}}{R^{2}}\right) 0 $\Rightarrow \int_{0}^{R} \rho_{0} 4 \pi\left(1-\frac{r^{2}}{R^{2}}\right) \cdot r^{2} d r=\frac{4}{3} \pi R^{3} \rho_{\ell}$ $\Rightarrow 4 \pi \rho_{0}\left[\frac{r^{3}}{3}-\frac{r^{5}}{5 R^{2}}\right]_{0}^{R}=\frac{4}{3} \pi R^{3} \rho_{\ell}$ $\frac{4 \pi \rho_{0} R^{3}}{3} \times \frac{2}{5}=\frac{4}{3} \pi R^{3} \rho_{\ell}$ $\therefore \rho_{\ell}=\frac{2 \rho_{0}}{5}$