First slide

Fluid statics

Question

A concrete sphere of radius R has a cavity of radius r which is packed with sawdust. The specific gravities of concrete and sawdust are respectively 2.4 and 0.3 for this sphere to float with its entire volume submerged under water. Ratio of mass of concrete to mass of sawdust will be

Difficult

A

8
B

4
C

3
D

zero

Solution

Let specific gravities of concrete and saw dust are $ρ_{1}$ and $ρ_{2}$ respectively.
According to principle of floatation weight of whole sphere = upthrust on the sphere
$\frac{4}{3} π (R^{3} - r^{3}) ρ_{1} g + \frac{4}{3} {πr}^{3} ρ_{2} g = \frac{4}{3} {πR}^{3} \times 1 \times g$
$\Rightarrow R^{3} ρ_{1} - r^{3} ρ_{1} + r^{3} ρ_{2} = R^{3}$
$\Rightarrow R^{3} (ρ_{1} - 1) = r^{3} (ρ_{1} - ρ_{2}) \Rightarrow \frac{R^{3}}{r^{3}} = \frac{ρ_{1} - ρ_{2}}{ρ_{1} - 1}$
$\Rightarrow \frac{R^{3} - r^{3}}{r^{3}} = \frac{ρ_{1} - ρ_{2} - ρ_{1} + 1}{ρ_{1} - 1}$
$\Rightarrow \frac{(R^{3} - r^{3}) ρ_{1}}{r^{3} ρ_{2}} = (\frac{1 - ρ_{2}}{ρ_{1} - 1}) \frac{ρ_{1}}{ρ_{2}}$
$\Rightarrow \frac{Mass of concrete}{Mass of saw dust} = (\frac{1 - 0.3}{2.4 - 1}) \times \frac{2.4}{0.3} = 4$

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