First slide

Pressure in a fluid-mechanical properties of fluids

Question

A hemispherical bowl just floats without sinking in a liquid of density 1.2 × 10³ kg/m³. If outer diameter and the density of the bowl are 1 m and 2 × 10⁴ kg/m³ respectively, then the inner diameter of the bowl will be

Moderate

A

0.94 m
B

0.90 m
C

0.98 m
D

0.99 m

Solution

Weight of the bowl = mg
$= Vρg = \frac{4}{3} π [{(\frac{D}{2})}^{3} - {(\frac{d}{2})}^{3}] ρg$
where D = Outer diameter,
d = Inner diameter
$ρ$ = Density of bowl
Weight of the liquid displaced by the bowl
$= Vσg = \frac{4}{3} π {(\frac{D}{2})}^{3} σ g$
where $σ$ is the density of the liquid.
For the flotation $\frac{4}{3} π {(\frac{D}{2})}^{3} σ g = \frac{4}{3} π [{(\frac{D}{2})}^{3} - {(\frac{d}{2})}^{3}] ρ g$
$\Rightarrow {(\frac{1}{2})}^{3} \times 1 .2 \times 10^{3} = [{(\frac{1}{2})}^{3} - {(\frac{d}{2})}^{3}] 2 \times 10^{4}$
By solving we get d = 0.98 m.

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