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Reynold's Number

Two spherical...

Question

Two spherical bodies having masses $m_{1}, m_{2}$ respectively falling in viscous medium the ratio of terminal velocities is $\frac{v_{1}}{v_{2}} = (\frac{m_{1}}{m_{2}})^{2 / 3}$ .

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Solution

Formula for terminal velocity (v) in a viscous medium ( $η$ ) of density ( $σ$ ) is given by:-
$v = \frac{2 r^{2}}{9 η} (P - σ)$
now mass = volume $\times$ density
for a spherical body $v = \frac{4}{3} π r^{3}$ , So,
$m = \frac{4}{3} π P r^{3}$
$= r^{2} = {(\frac{3 m}{4 π p})}^{2 / 3}$ .....2
Hence ratio form $1$ and $2$
$\frac{V_{1}}{V_{2}} = (\frac{r_{1}}{r_{2}}) = {(\frac{m_{1}}{m_{2}})}^{2 / 3}$

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