Question

A solid sphere of density $\rho ={\rho }_0(1-\frac{r^2}{R^2}),0<r\le R$ (where $r$ is the distance from the centre of the sphere) just floats in a liquid. Then, density of the liquid is -

• A. $\frac{2{\rho }_o}{5}$
• B. $\frac{4{\rho }_o}{5}$
• C. $\frac{2{\rho }_o}{7}$
• D. ${\rho }_0$

Answer 1

The correct option is A $\frac{2{\rho }_o}{5}$

From the data given in the question,
$\rho ={\rho }_0(1-\frac{r^2}{R^2}),0<r\le R$
Condition for floatation -
Weight of solid sphere = Force of buoyancy = Weight of liquid displaced
i.e ${\int }_0^R\rho \left(4\pi r^{2}dr\right)g={\rho }_L{V}_{L}g$
$\Rightarrow {\int }_0^R\rho \left(4\pi r^{2}dr\right)={\rho }_L\frac{4}{3}\pi R^3$
$\Rightarrow {\int }_0^R{\rho }_0(1-\frac{r^2}{R^2})4\pi r^{2}dr={\rho }_L\frac{4}{3}\pi R^3$
[substituting relation given in question]
$\Rightarrow {\int }_0^R{\rho }_{0}4\pi (r^2-\frac{r^4}{R^2})dr={\rho }_{0}4\pi {(\frac{r^3}{3}-\frac{r^5}{5R^2})}_0^R={\rho }_L\frac{4}{3}\pi R^3$
$\Rightarrow \frac{8{\rho }_0\pi R^3}{15}={\rho }_L\frac{4}{3}\pi R^3$
$\Rightarrow {\rho }_L=\frac{2}{5}{\rho }_0$
Thus, option (a) is the correct answer.