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Two spherical bodies having masses m_{1},m_{2} respectively fa

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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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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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Text solutionVerified

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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Question Text	Two spherical bodies having masses $m_{1}, m_{2}$ respectively falling in viscous medium the ratio of terminal velocities is $\frac{v _{1}}{v _{2}} = (\frac{m _{1}}{m _{2}})^{2/3}$ .
Answer Type	Text solution:1
Upvotes	150