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(B) Let the specific gravities of concrete and sawdust be $\rho_1 = 2.4$ and $\rho_2 = 0.3$ respectively.According to the principle of floatation,the

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$4$

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(B) Let the specific gravities of concrete and sawdust be $ho_1 = 2.4$ and $ho_2 = 0.3$ respectively.According to the principle of floatation,the weight of the whole sphere equals the upthrust (buoyant force) on the sphere.Weight of sphere = Weight of water displaced$\frac{4}{3}\pi (R^3 - r^3)ho_1 g + \frac{4}{3}\pi r^3 ho_2 g = \frac{4}{3}\pi R^3 	imes 1 	imes g$Dividing by $\frac{4}{3}\pi g$,we get:$(R^3 - r^3)ho_1 + r^3 ho_2 = R^3$$R^3 ho_1 - r^3 ho_1 + r^3 ho_2 = R^3$$R^3(ho_1 - 1) = r^3(ho_1 - ho_2)$$\frac{R^3}{r^3} = \frac{ho_1 - ho_2}{ho_1 - 1} = \frac{2.4 - 0.3}{2.4 - 1} = \frac{2.1}{1.4} = 1.5$Now,the ratio of the mass of concrete $(M_c)$ to the mass of sawdust $(M_s)$ is:$\frac{M_c}{M_s} = \frac{\frac{4}{3}\pi (R^3 - r^3)ho_1}{\frac{4}{3}\pi r^3 ho_2} = \left( \frac{R^3}{r^3} - 1 ight) \frac{ho_1}{ho_2}$Substituting the values:$\frac{M_c}{M_s} = (1.5 - 1) 	imes \frac{2.4}{0.3} = 0.5 	imes 8 = 4$.

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