Explanation

(b) 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
$\frac{4}{3}\pi\left( R^{3} - r^{3} \right)\rho_{1}g + \frac{4}{3}\pi r^{3}\ \rho_{2}g = \frac{4}{3}\pi R^{3} \times 1 \times g$
$\Rightarrow R^{3}\rho_{1} - r^{3}\rho_{1} + r^{3}\rho_{2} = R^{3}$
$\Rightarrow R^{3}\left( \rho_{1} - 1 \right) = r^{3}\left( \rho_{1} - \rho_{2} \right) \Rightarrow \frac{R^{3}}{r^{3}} = \frac{\rho_{1} - \rho_{2}}{\rho_{1} - 1}$
$\Rightarrow \frac{R^{3} - r^{3}}{r^{3}} = \frac{\rho_{1} - \rho_{2} - \rho_{1} + 1}{\rho_{1} - 1}$
$\Rightarrow \frac{{(R}^{3} - r^{3})\rho_{1}}{r^{3}\rho_{2}} = \left( \frac{{1 - \rho}_{2}}{\rho_{1} - 1} \right)\frac{\rho_{1}}{\rho_{2}}$
$\Rightarrow \ \frac{Mass\ of\ concrete}{Mass\ of\ saw\ dust} = \left( \frac{1 - 0.3}{2.4 - 1} \right) \times \frac{2.4}{0.3} = 4$