Q 11-09-130JEE MainJEE Main 2020 (8 Jan, Shift 1)Medium
Consider a solid sphere of radius $R$ and mass density $\rho(r) = \rho_0\left(1 - \dfrac{r^2}{R^2}\right)$, $0 < r \le R$. The minimum density of a liquid in which it will float is:
Answer: (C) $\dfrac{2\rho_0}{5}$
$$M = \int_0^R \rho_0\left(1 - \frac{r^2}{R^2}\right)4\pi r^2\,dr = 4\pi\rho_0\left(\frac{R^3}{3} - \frac{R^3}{5}\right) = \frac{8\pi\rho_0R^3}{15}$$
Average density:
$$\bar\rho = \frac{M}{\frac43\pi R^3} = \frac{8}{15}\cdot\frac34\rho_0 = \frac{2\rho_0}{5}$$
The sphere floats if the liquid is at least this dense, so the minimum density is $\dfrac{2\rho_0}{5}$.
Solution by Sreeraj P, M.Sc Physics