Consider a solid sphere of radius $R$ and mass density
Question:

Consider a solid sphere of radius $R$ and mass density $\rho(r)=\rho_{0}\left(1-\frac{r^{2}}{R^{2}}\right), 0<r \leq R .$ The minimum density of a

liquid in which it will float is:

1. (1) $\frac{\rho_{0}}{3}$

2. (2) $\frac{\rho_{0}}{5}$

3. (3) $\frac{2 \rho_{0}}{5}$

4. (4) $\frac{2 \rho_{0}}{3}$

Correct Option: , 3

Solution:

(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 $\int \rho d V=\frac{4}{3} \pi R^{3} \rho_{\ell}$

$\left[\rho(r)=\rho_{0}\left(1-\frac{r^{2}}{R^{2}}\right) 0<r \leq R\right.$ given $]$

$\Rightarrow \int_{0}^{R} \rho_{0} 4 \pi\left(1-\frac{r^{2}}{R^{2}}\right) \cdot r^{2} d r=\frac{4}{3} \pi R^{3} \rho_{\ell}$

$\Rightarrow 4 \pi \rho_{0}\left[\frac{r^{3}}{3}-\frac{r^{5}}{5 R^{2}}\right]_{0}^{R}=\frac{4}{3} \pi R^{3} \rho_{\ell}$

$\frac{4 \pi \rho_{0} R^{3}}{3} \times \frac{2}{5}=\frac{4}{3} \pi R^{3} \rho_{\ell}$

$\therefore \rho_{\ell}=\frac{2 \rho_{0}}{5}$