Question

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

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

Answer 1

The correct option is A

$\left(\frac{2}{5}\right){\rho }_0$

Step 1. Given Data,

A solid sphere of density $=\rho \left(r\right)$$={\rho }_0\left(1-\frac{r^2}{R^2}\right),0<r\le R$

Where ${\rho }_0$ is the density of the liquid $r,R$ are radius.

The minimum density of a liquid $={\rho }_l$

Step 2. Find out the minimum density of a liquid ${\rho }_l$ in which it floats,

Let the mass of the sphere be $m$ and the density of the liquid be ${\rho }_0$

$\rho \left(r\right)={\rho }_0\left(1-\frac{r^2}{R^2}\right),0<r\le R$

(Note: Since, the sphere is floating in the liquid, Buoyancy force ($F_B$) due to liquid will balance the weight of the sphere.)

$$\begin{gathered} W=mg \\ \Rightarrow {\rho }_l\left(\frac{4\mathrm{\pi }}{3}\right)R^{3}g=\int \rho \left(4{\mathrm{\pi r}}^2\mathrm{dr}\right)g \\ \Rightarrow {\rho }_l\left(\frac{4\mathrm{\pi }}{3}\right)R^3=\int {\rho }_0\left(1-\frac{r^2}{R^2}\right)\left(4{\mathrm{\pi r}}^2\mathrm{dr}\right) \\ \Rightarrow {\rho }_l\left(\frac{4\mathrm{\pi }}{3}\right)R^3={\int }_0^R{\rho }_0\left(1-\frac{r^2}{R^2}\right)\left(4{\mathrm{\pi r}}^2\mathrm{dr}\right) \\ \Rightarrow {\rho }_l\left(\frac{4\mathrm{\pi }}{3}\right)R^3=4\mathrm{\pi }{\rho }_0{\left[\frac{r^3}{3}-\frac{r^5}{5R^2}\right]}_0^R \\ \Rightarrow {\rho }_l=\left(\frac{2}{5}\right){\rho }_0 \end{gathered}$$

Hence, the correct option is $\text{'}A\text{'}$.