First slide

Fluid statics

Question

A concrete sphere of radius fl has cavity of radius r which is packed with saw dust. The specific gravity of concrete and saw dust 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 saw dust will be

Moderate

A

8
B

4
C

3
D

0

Solution

The weight of whole sphere will be equal to the upthrust' Hence
$\frac{4}{3} π (R^{3} - r^{3}) ρ_{1} + \frac{4}{3} {πr}^{3} ρ_{2} = \frac{4}{3} {πR}^{3} \times 1 where p_{1} and p_{2} are the specific Sravities of concrete and saw dust respectivelY R^{3} ρ_{1} - r^{3} ρ_{1} + r^{3} ρ_{2} = R^{3} or R^{3} (ρ_{1} - 1) = r^{3} (ρ_{1} - ρ_{2}) or \frac{R^{3}}{r^{3}} = \frac{(ρ_{1} - ρ_{2})}{ρ_{1} - 1} or \frac{R^{3} - r^{3}}{r^{3}} = \frac{ρ_{1} - ρ_{2} - ρ_{1} + 1}{ρ_{1} - 1} or \frac{(R^{3} - r^{3}) ρ_{1}}{r^{3} ρ_{2}} = (\frac{1 - ρ_{2}}{ρ_{1} - 1}) \times \frac{ρ_{1}}{ρ_{2}} or \frac{Mass of concrete}{Mass of saw dust} = \frac{1 - 0 \cdot 3}{2 \cdot 4 - 1} \times \frac{2 \cdot 4}{0 \cdot 3} = 4$

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