Question

A concrete sphere of radius $R$ has a cavity of radius $r$ which is packed with sawdust. The specific gravities of concrete and sawdust are respectively $2.4$ and $0.3$ for this sphere to float with its entire volume submerged under water. What is the ratio of mass of concrete to mass of sawdust?

Answer 1

Let specific gravities of concrete and saw dust are $\rho_{1}$, and $\rho_{2}$, respectively. According to principle of floatation weight of whole sphere= upthrust on the sphere
$\frac{4}{3}\pi\left(R^{3} -r^{3}\right) \rho_{1}g+\frac{4}{3}\pi r^{3}\rho_{2}g = \frac{4}{3}\pi R^{3} \times1\times g $
$\Rightarrow R^{3}\rho_{1} -r^{3}\rho_{1} +r^{3} \rho_{2} =R^{3}$
$\Rightarrow R^{3}\left(\rho_{1} -1\right) =r^{3}\left(\rho_{1} -\rho_{1}\right)$
$\Rightarrow \frac{R^{3}}{r^{3}} =\frac{\rho_{1} -\rho_{2}}{\rho_{1} -1}$
$\Rightarrow \frac{R^{3} -r^{3}}{r^{3}} =\frac{\rho_{1} -\rho_{2} -\rho_{1} +1}{\rho_{1} -1}$
$\Rightarrow \frac{\left(R^{3} -r^{3}\right)\rho_{1}}{r^{3} \rho_{2}} = \left(\frac{1-\rho_{2}}{\rho_{1}-1}\right) \frac{\rho_{1}}{\rho_{2}}$
$\Rightarrow \frac{\text{Mass of concrete}}{\text{Mass of saw dust}} =\left(\frac{1 -0.3}{2.4-1}\right) \times \frac{2.4}{0.3} = 4$