Question

A concrete sphere of radius $R$ has a cavity of radius $r$ which is placed 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 1

Answer: (B)

$4$

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

$\frac{4}{3}\pi (R^3-r^3){\rho }_{1}g+\frac{4}{3}\pi r^3{\rho }_{2}g=\frac{4}{3}\pi R^3\times 1\times g$

$\Rightarrow R^3{\rho }_1-r^3{\rho }_1+r^3{\rho }_2=R^3$

$\Rightarrow R^3({\rho }_1-1)=r^3({\rho }_1-{\rho }_2)\Rightarrow \frac{R^3}{r^3}=\frac{{\rho }_1-{\rho }_2}{{\rho }_1-1}$

$\Rightarrow \frac{R^3-r^3}{r^3}=\frac{{\rho }_1-{\rho }_2-{\rho }_1+1}{{\rho }_1-1}$

$\Rightarrow \frac{(R^3-r^3){\rho }_1}{r^3{\rho }_2}=\left(\frac{1-{\rho }_2}{{\rho }_1-1}\right)\frac{{\rho }_1}{{\rho }_2}$

$\Rightarrow \frac{Massofconcrete}{Massofsawdust}=\left(\frac{1-0.3}{2.4-1}\right)\times \frac{2.4}{0.3}=4$