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(C) The terminal velocity $v$ of an air bubble rising in a liquid is given by the formula: $v = \frac{2}{9} \frac{r^2 \rho g}{\eta}$,where $\rho$ is t

Vedclass · Accepted Answer

$130$

Vedclass · Answer

(C) The terminal velocity $v$ of an air bubble rising in a liquid is given by the formula: $v = \frac{2}{9} \frac{r^2 ho g}{\eta}$,where $ho$ is the density of the liquid,$r$ is the radius,and $\eta$ is the coefficient of viscosity.Rearranging for $\eta$: $\eta = \frac{2}{9} \frac{r^2 ho g}{v}$.Given values: $r = 1 \ cm = 10^{-2} \ m$,$ho = 1.5 \ g/cc = 1.5 	imes 10^3 \ kg/m^3$,$v = 0.25 \ cm/s = 0.25 	imes 10^{-2} \ m/s$,and $g = 9.8 \ m/s^2$.Substituting the values: $\eta = \frac{2}{9} \cdot \frac{(10^{-2})^2 \cdot (1.5 	imes 10^3) \cdot 9.8}{0.25 	imes 10^{-2}}$.$\eta = \frac{2}{9} \cdot \frac{10^{-4} \cdot 1500 \cdot 9.8}{0.0025} = \frac{2}{9} \cdot \frac{0.15 \cdot 9.8}{0.0025} = \frac{2}{9} \cdot \frac{1.47}{0.0025} = \frac{2}{9} \cdot 588 \approx 130.6 \ Pa \cdot s$.Thus,the coefficient of viscosity is approximately $130 \ Pa \cdot s$.

An air bubble of radius $1 \ cm$ rises from the bottom portion through a liquid of density $1.5 \ g/cc$ at a constant speed of $0.25 \ cm/s$. If the density of air is neglected,the coefficient of viscosity of the liquid is approximately,(in $Pa \cdot s$):

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