Question

For a concrete sphere of radius $R$ having a cavity of radius $r$ packed with sawdust, to float with its entire volume submerged under water what will be the ratio of mass of concrete to mass of sawdust? Given the specific gravities of concrete and sawdust are respectively $2.4$ and $0.3$ .

• A. $8$
• B. $4$
• C. $3$
• D. zero

Answer 1

Answer: (B)

Let specific gravities of concrete and sawdust are $\rho _{1}$ and $\rho _{2}$ respectively.
According to principle of floatation, weight of whole sphere = upthrust on the sphere
$\frac{4}{3} \pi \left(\left(\text{R}\right)^{3} - \left(\text{r}\right)^{3}\right) \left(\rho \right)_{1} \text{g} + \frac{4}{3} \pi \left(\text{r}\right)^{3} \left( \rho \right)_{2} \text{g} = \frac{4}{3} \pi \left(\text{R}\right)^{3} \times 1 \times \text{g}$
$\Rightarrow \text{R}^{3} \rho _{1} - \text{r}^{3} \rho _{1} + \text{r}^{3} \rho _{2} = \text{R}^{3}$
$\Rightarrow \left(\text{R}\right)^{3} \left(\left(\rho \right)_{1} - 1\right) = \left(\text{r}\right)^{3} \left(\left(\rho \right)_{1} - \left(\rho \right)_{2}\right) \Rightarrow \frac{\left(\text{R}\right)^{3}}{\left(\text{r}\right)^{3}} = \frac{\left(\rho \right)_{1} - \left(\rho \right)_{2}}{\left(\rho \right)_{1} - 1}$
$\Rightarrow \frac{\left(\text{R}\right)^{3} - \left(\text{r}\right)^{3}}{\left(\text{r}\right)^{3}} = \frac{\left(\rho \right)_{1} - \left(\rho \right)_{2} - \left(\rho \right)_{1} + 1}{\left(\rho \right)_{1} - 1} \Rightarrow \frac{\left(\left(\text{R}\right)^{3} - \left(\text{r}\right)^{3}\right) \left(\rho \right)_{1}}{\left(\text{r}\right)^{3} \left(\rho \right)_{2}} = \left(\right. \frac{1 - \left(\rho \right)_{2}}{\left(\rho \right)_{1} - 1} \left.\right) \frac{\left(\rho \right)_{1}}{\left(\rho \right)_{2}}$
$\Rightarrow \frac{\text{Mass of concrete}}{\text{Mass of saw dust}} = \left(\right. \frac{1 - 0 \text{.} 3}{2 \text{.} 4 - 1} \left.\right) \times \frac{2 \text{.} 4}{\text{0} \text{.} 3} = 4$