Question

A concrete sphere of radius $R$ has cavity of radius $r$ which is packed with saw-dust. The specific gravities of concrete and sawdust are $2.4$ and $0.3$ respectively. The sphere floats with its entire volume sub-merged under water. Calculate the ratio of mass of concrete and the mass of saw-dust.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

The specific gravities of concrete and sawdust are ${\rho }_1$ and ${\rho }_2$ respectively.

Given that,

${\rho }_1=2.4$

${\rho }_2=0.3$

According to the flotation,

Weight of whole sphere = upthrust on the sphere

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

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

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

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

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

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

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

$\frac{masssofconcrete}{massofsawdust}=\frac{1-0.3}{2.4-1}\times \frac{2.4}{0.3}$

$\frac{masssofconcrete}{massofsawdust}=4$

Hence, the ratio of mass of concrete and the mass of saw-dust is $4$.