First slide

Fluid statics

Question

A hemispherical bowl just floats without sinking in a liquid of density $1.2 \times 10^{3} kg / m^{3}$ . If outer diameter and the density of the bowl are I m and 2 x $l 0^{4} kg / m^{3}$ respectively, then the inner diameter of the bowl will be

Difficult

A

0.94 m
B

0.97 m
C

0.98 m
D

0.99 m

Solution

Weight of the bowl = mg
= $V ρ g = \frac{4}{3} π [{(\frac{D}{2})}^{3} - {(\frac{d}{2})}^{3}] ρg$
Where D = Outer diameter
d = Inner diameter
$ρ$ = Density of bowl
Weight of the liquid dispalced by the bowl
= $V σ g = \frac{4}{3} π {(\frac{D}{2})}^{3} σg$
where $σ$ is the density of the liquid.
For the flotation
$\frac{4}{3} π {(\frac{D}{2})}^{3} σg = \frac{4}{3} π [{(\frac{D}{2})}^{3} - {(\frac{d}{2})}^{3}] ρg$
$\Rightarrow {(\frac{1}{2})}^{3} \times 1.2 \times 10^{3} = [{(\frac{1}{2})}^{3} - {(\frac{d}{2})}^{3}] 2 \times 10^{4}$
By solving we get d = 0.98 m.

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