Question

A gas bubble of $ 2 \,cm $ diameter rises through a liquid of density $ 1.75 \,gm \,cm^{-3} $ with a fixed speed of $ 0.35 \,cm\, s^{-1} $ . Neglect the density of the gas. The coefficient of viscosity of the liquid is

• A. $ 870 $ poise
• B. $ 1120 $ poise
• C. $ 982 $ poise
• D. $ 1089 $ poise

Answer 1

Answer: (D)

As gas bubble is moving upward with constant speed so Buoyant force = Viscous drag force
$ \Rightarrow F_{b}=F_{v} \Rightarrow \left(\frac{4}{3}\pi r^{3}\right) \rho g=6\pi\eta rv $

$ \Rightarrow \eta=\frac{2}{9} \frac{r^{2}\rho g}{v} $
Here, $ r= 1 cm, \rho=1.75 \, g \, cm^{-3} $ ,
$ g = 980\, cm \, s^{-2} $
$ v = 0.35 \,cm \, s^{-1}, \eta = $ ?
$ \therefore \eta =\frac{2}{9}\times\frac{1^{2}\times1.75\times980}{0.35} $
$ =1088.88 \approx 1089 $ poise