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

Answer: 4

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
$\frac{4}{3}\pi \left(R^3-r^3\right){\rho }_{1}g+\frac{4}{3}\pi r^3{\rho }_{2}g=\frac{4}{3}\pi R^3\times 1\times g$
$\Rightarrow R^3{\rho }_1-r^3{\rho }_1+r^3{\rho }_2=R^3$
$\Rightarrow R^3\left({\rho }_1-1\right)=r^3\left({\rho }_1-{\rho }_1\right)$
$\Rightarrow \frac{R^3}{r^3}=\frac{{\rho }_1-{\rho }_2}{{\rho }_1-1}$
$\Rightarrow \frac{R^3-r^3}{r^3}=\frac{{\rho }_1-{\rho }_2-{\rho }_1+1}{{\rho }_1-1}$
$\Rightarrow \frac{\left(R^3-r^3\right){\rho }_1}{r^3{\rho }_2}=\left(\frac{1-{\rho }_2}{{\rho }_1-1}\right)\frac{{\rho }_1}{{\rho }_2}$
$\Rightarrow \frac{\text{Mass of concrete}}{\text{Mass of saw dust}}=\left(\frac{1-0.3}{2.4-1}\right)\times \frac{2.4}{0.3}=4$