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. Ratio of mass of concrete to mass of sawdust will be
A.
8
B.
4
C.
3
D.
zero
Answer :
4
Solution :
$$\eqalign{ & \left( {{m_{{\text{concrete}}}} + {m_{{\text{sawdust}}}}} \right)g = V{\rho _w}g \cr & 2.4 \times \frac{4}{3}\pi \left( {{R^3} - {r^3}} \right) + 0.3 \times \frac{4}{3}\pi {r^3} = \frac{4}{3}\pi {R^3} \times 1 \cr & {\text{or}}\,\,{R^3} = \frac{3}{2}{r^3} \cr & \therefore \frac{{{m_{{\text{concrete}}}}}}{{{m_{{\text{sawdust}}}}}} = \frac{{{\rho _{{\text{concrete}}}}\frac{4}{3}\pi \left( {{R^3} - {r^3}} \right)}}{{{\rho _{{\text{sawdust}}}}\frac{4}{3}\pi {r^3}}} \cr & = \frac{{2.4}}{{0.3}}\frac{{\left( {\frac{3}{2}{r^3} - {r^3}} \right)}}{{{r^3}}} = 4 \cr} $$
$$\eqalign{ & \left( {{m_{{\text{concrete}}}} + {m_{{\text{sawdust}}}}} \right)g = V{\rho _w}g \cr & 2.4 \times \frac{4}{3}\pi \left( {{R^3} - {r^3}} \right) + 0.3 \times \frac{4}{3}\pi {r^3} = \frac{4}{3}\pi {R^3} \times 1 \cr & {\text{or}}\,\,{R^3} = \frac{3}{2}{r^3} \cr & \therefore \frac{{{m_{{\text{concrete}}}}}}{{{m_{{\text{sawdust}}}}}} = \frac{{{\rho _{{\text{concrete}}}}\frac{4}{3}\pi \left( {{R^3} - {r^3}} \right)}}{{{\rho _{{\text{sawdust}}}}\frac{4}{3}\pi {r^3}}} \cr & = \frac{{2.4}}{{0.3}}\frac{{\left( {\frac{3}{2}{r^3} - {r^3}} \right)}}{{{r^3}}} = 4 \cr} $$
