Question

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

• A. $\frac{{\rho }_0}{3}$
• B. $\frac{{\rho }_0}{5}$
• C. $\frac{2{\rho }_0}{5}$
• D. $\frac{2{\rho }_0}{3}$

Answer 1

The correct option is C $\frac{2{\rho }_0}{5}$

For minimum density of liquid, the solid sphere has to float (completely immersed) in the liquid, and it will happen when,

$mg=F_B(\text{also}{V}_{\text{immersed}}=V_{\text{total}})$
$\Rightarrow mg={\rho }_{l}Vg$
$\Rightarrow \int \rho dV=\frac{4}{3}\pi R^3{\rho }_l$

Given that,
$\rho \left(r\right)={\rho }_0(1-\frac{r^2}{R^2})\text{for}0<r\le R$
$V=\frac{4}{3}\pi r^3$
$\Rightarrow dV=4\pi r^{2}dr$

${\int }_0^R{\rho }_0(1-\frac{r^2}{R^2}).4\pi r^{2}dr=\frac{4}{3}\pi R^3{\rho }_l$
$\Rightarrow 4\pi {\rho }_0{[\frac{r^3}{3}-\frac{r^5}{5R^2}]}_0^R=\frac{4}{3}\pi R^3{\rho }_l$
$\frac{4\pi {\rho }_0{R}^3}{3}\times \frac{2}{5}=\frac{4}{3}\pi R^3{\rho }_l$
$\therefore {\rho }_l=\frac{2{\rho }_0}{5}$

Hence, option $\left(C\right)$ is correct.