Answer: (3)

Solution

For minimum density of liquid, solid sphere has to float (completely immersed) in the liquid. $mg = F_B$ (also $V_{\text{immersed}} = V_{\text{total}}$) or $\int \rho dV = \frac{4}{3}\pi R^3 \rho_l$ $\left[\rho(r) = \rho_0\left(1 - \frac{r^2}{R^2}\right), 0 < r \leq R \text{ given}\right]$ $\Rightarrow \int_0^R \rho_0 4\pi\left(1 - \frac{r^2}{R^2}\right)r^2 dr = \frac{4}{3}\pi R^3 \rho_l$ $\Rightarrow 4\pi\rho_0\left(\frac{r^3}{3} - \frac{r^5}{5R^2}\right)_0^R = \frac{4}{3}\pi R^3 \rho_l$ $\frac{4\pi\rho_0 R^3}{3} \times \frac{2}{5} = \frac{4}{3}\pi R^3 \rho_l$ $\therefore \rho_l = \frac{2\rho_0}{5}$