Question

A concrete sphere of radius $R$ has a 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 submerged under water. Calculate the ratio of the mass of concrete and the mass of saw-dust.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D

$4$

Step 1: Given data

Radius of sphere(concrete) = $R$

Radius of cavity(saw-dust) = $r$

Specific gravity(concrete)= ${\rho }_1$ = $2.4$

Specific gravity(saw-dust) = ${\rho }_2$ = $0.3$

Step 2: Formula used

According to the principle of floatation, the weight of a floating body is equal to the weight of the liquid displaced by the submerged part.

Therefore,

Weight of the whole sphere = buoyant force on the sphere

Weight of the concrete sphere + weight of the cavity containing sawdust = buoyant force on the sphere

Weight of the concrete sphere$=\mathrm{Mg}={\mathrm{V\rho }}_1\mathrm{g}=\frac{4}{3}\mathrm{\pi }({\mathrm{R}}^3-{\mathrm{r}}^3){\mathrm{\rho }}_1\mathrm{g}$

Weight of the cavity containing sawdust$=\mathrm{mg}={\mathrm{v\rho }}_2\mathrm{g}=\frac{4}{3}\mathrm{\pi }{\mathrm{r}}^3{\mathrm{\rho }}_2\mathrm{g}$

The buoyant force on the sphere$=\mathrm{m}\text{'}\mathrm{g}=\mathrm{V\rho g}=\frac{4}{3}\mathrm{\pi }{\mathrm{R}}^3\mathrm{\rho g}$

$$\begin{gathered} \frac{4}{3}\mathrm{\pi }({\mathrm{R}}^3-{\mathrm{r}}^3){\mathrm{\rho }}_1\mathrm{g}+\frac{4}{3}\mathrm{\pi }{\mathrm{r}}^3{\mathrm{\rho }}_2\mathrm{g}=\frac{4}{3}\mathrm{\pi }{\mathrm{R}}^3\mathrm{\rho g}(\because \mathrm{\rho }=1(\mathrm{water}\left)\right) \\ \Rightarrow ({\mathrm{R}}^3-{\mathrm{r}}^3){\mathrm{\rho }}_1+{\mathrm{r}}^3{\mathrm{\rho }}_2={\mathrm{R}}^3 \\ \Rightarrow {\mathrm{R}}^3{\mathrm{\rho }}_1-{\mathrm{r}}^3\left({\mathrm{\rho }}_1-{\mathrm{\rho }}_2\right)={\mathrm{R}}^3 \\ \Rightarrow \frac{{\mathrm{R}}^3}{{\mathrm{r}}^3}=\frac{{\mathrm{\rho }}_1-{\mathrm{\rho }}_2}{{\mathrm{\rho }}_1-1} \\ \Rightarrow \frac{{\mathrm{R}}^3}{{\mathrm{r}}^3}-1=\frac{{\mathrm{\rho }}_1-{\mathrm{\rho }}_2}{{\mathrm{\rho }}_1-1}-1 \\ \Rightarrow \frac{{\mathrm{R}}^3-{\mathrm{r}}^3}{{\mathrm{r}}^3}=\frac{{\mathrm{\rho }}_1-{\mathrm{\rho }}_2-{\mathrm{\rho }}_1+1}{{\mathrm{\rho }}_1-1} \\ \Rightarrow \frac{({\mathrm{R}}^3-{\mathrm{r}}^3){\mathrm{\rho }}_1}{{\mathrm{r}}^3{\mathrm{\rho }}_2}=\left(\frac{1-{\mathrm{\rho }}_2}{{\mathrm{\rho }}_1-1}\right)\frac{{\mathrm{\rho }}_1}{{\mathrm{\rho }}_2} \end{gathered}$$

Multiplying numerator and denominator of the LHS by $\frac{4}{3}\mathrm{\pi }$ we get,

$$\begin{gathered} \frac{{\displaystyle \frac{4}{3}\mathrm{\pi }}({\mathrm{R}}^3-{\mathrm{r}}^3){\mathrm{\rho }}_1}{{\displaystyle \frac{4}{3}\mathrm{\pi }}{\mathrm{r}}^3{\mathrm{\rho }}_2}=\left(\frac{1-{\mathrm{\rho }}_2}{{\mathrm{\rho }}_1-1}\right)\frac{{\mathrm{\rho }}_1}{{\mathrm{\rho }}_2} \\ \Rightarrow \frac{\mathrm{mass}\mathrm{of}\mathrm{concrete}}{\mathrm{mass}\mathrm{of}\mathrm{sawdust}}=\left(\frac{1-{\mathrm{\rho }}_2}{{\mathrm{\rho }}_1-1}\right)\frac{{\mathrm{\rho }}_1}{{\mathrm{\rho }}_2} \end{gathered}$$

Step 3: Substitute values in the formula to get the final answer

$$\begin{gathered} \frac{\mathrm{mass}\mathrm{of}\mathrm{concrete}}{\mathrm{mass}\mathrm{of}\mathrm{sawdust}}=\left(\frac{1-{\mathrm{\rho }}_2}{{\mathrm{\rho }}_1-1}\right)\frac{{\mathrm{\rho }}_1}{{\mathrm{\rho }}_2} \\ \Rightarrow \frac{\mathrm{mass}\mathrm{of}\mathrm{concrete}}{\mathrm{mass}\mathrm{of}\mathrm{sawdust}}=\frac{1-0.3}{2.4-1}\times \frac{2.4}{0.3} \\ \Rightarrow \frac{\mathrm{mass}\mathrm{of}\mathrm{concrete}}{\mathrm{mass}\mathrm{of}\mathrm{sawdust}}=\frac{0.7}{1.4}\times 8 \\ \Rightarrow \frac{\mathrm{mass}\mathrm{of}\mathrm{concrete}}{\mathrm{mass}\mathrm{of}\mathrm{sawdust}}=4 \end{gathered}$$

The ratio of the mass of concrete and the mass of saw-dust is $4$

Therefore, the correct option is D.