Question

A concrete sphere of radius $R$ has a spherical 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, ratio of mass of concrete to the mass of sawdust will be

• A. $8$
• B. $4$
• C. $3$
• D. $1$

Answer 1

The correct option is B $4$


Let $V$ be the volume of the sphere and ${\rho }_w$ be the density of the water.
Since sphere is fully submerged under water, then
Weight of whole sphere = Weight of water displaced
$(m_{\text{concrete}}+m_{\text{sawdust}})g=V{\rho }_{w}g$
$\Rightarrow 2.4\times \frac{4}{3}\pi (R^3-r^3)+0.3\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\times 1$
or $R^3=\frac{3}{2}{r}^3$
$\therefore \frac{m_{concrete}}{m_{sawdust}}=\frac{\frac{4}{3}\pi (R^3-r^3){\rho }_{concrete}}{\frac{4}{3}\pi r^3{\rho }_{sawdust}}$
$\Rightarrow \frac{m_{concrete}}{m_{sawdust}}=\frac{2.4}{0.3}\frac{(\frac{3}{2}{r}^3-r^3)}{r^3}=4$