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A concrete sphere of radius R has a cavity of radius r which is packed with saw-dust. The specific gravities of concrete and saw-dust 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

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Detailed Solution

law of floatation

Weight of whole sphere = up thrust on the sphere

$\frac{4}{3} π (R^{3} - r^{3}) ρ_{1} g + \frac{4}{3} π r^{3} ρ_{2} g$

$= \frac{4}{3} π R^{3} \times 1 \times g$

$\frac{R^{3}}{r^{3}} = \frac{ρ_{1} - ρ_{2}}{ρ_{1} - 1} \Rightarrow \frac{R^{3} - r^{3}}{r^{3}} = \frac{1 - ρ_{2}}{ρ_{1} - 1}$

$\frac{R^{3} - r^{3}}{r^{3}} . \frac{ρ_{1}}{ρ_{2}} = \frac{1 - ρ_{2}}{ρ_{1} - 1} \times \frac{ρ_{1}}{ρ_{2}}$

$\frac{mass of concrete}{mass of dust} = \frac{(1 - 0.3)}{(2.4 - 1)} \times \frac{2.4}{0.3} = 4$

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