With what the terminal velocity will an air bubble $1\;mm$ in diameter rise in a liquid of viscosity $150\;poise$ and density $0.9\;g/cm^3$?
Given,
Radius of the air bubble $(r) = 0.5\;mm = 0.5 * 10^{-3}\;m$
Viscosity of the oil $(\eta) = 150 \;Poise = 15\;decapoise$
Density of the air $(\rho) = 0$
Density of the oil $(\sigma) = 0.9 * 1000 = 900 \;kg/m^3$
Terminal velocity $(v) = ?$
We have,
The $-\;ve$ sign indicates that the ball moves in the upward direction against the force of gravity.
Radius of the air bubble $(r) = 0.5\;mm = 0.5 * 10^{-3}\;m$
Viscosity of the oil $(\eta) = 150 \;Poise = 15\;decapoise$
Density of the air $(\rho) = 0$
Density of the oil $(\sigma) = 0.9 * 1000 = 900 \;kg/m^3$
Terminal velocity $(v) = ?$
We have,
$v = $ $\frac{2\;r^2 (\rho \; - \; \sigma)\,g}{9\;\eta} = \frac{2 \; * \; (0.5 \; * \; 10^{-3})^2 \; * \;(0 \; - \; 900 ) \; * \; 9.8}{9 \; * \; 15}$
$ = - 3.27 * 10^{-5} \; m/s$
The $-\;ve$ sign indicates that the ball moves in the upward direction against the force of gravity.
Return to Main Menu
Comments
Post a Comment