You visited us 1 times! Enjoying our articles? Unlock Full Access!

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 $\frac{v_{1}}{v_{2}} = (\frac{m_{1}}{m_{2}})^{2 / 3}$ .

Open in App

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 $2$
$\frac{V_{1}}{V_{2}} = (\frac{r_{1}}{r_{2}}) = {(\frac{m_{1}}{m_{2}})}^{2 / 3}$

Suggest Corrections

Similar questions

Two bodies of masses m₁ and m₂ having equal kinetic energies. If p₁ and p₂ are their momentum respectively, Then the ratio of p₁ and p₂ is.

M_{1}

M_{2}

(M_{1} = \frac{M_{2}}{2})

v_{1}

v_{2}

v_{1} = 2 v_{2}

m_{1}

m_{2}

v_{1}

v_{2}

m_{1}

m_{2}

v_{1}

v_{2}

t = 0

t_{0}

{v^{'}}_{1}

{v^{'}}_{2}

2 t_{0}

[(m_{1} {v^{'}}_{1} + m_{2} {v^{'}}_{2}) - (m_{1} v_{1} + m_{2} v_{2})]

m_{1}

m_{2}

m_{2}

m_{1}

v_{1}

v_{2}

Join BYJU'S Learning Program