An air bubble of radius 1cm rises from the bottom portion through a liquid of density 1.5g/cc at constant speed of 0.25cm/se. If the density of air is neglected the coefficient of viscosity of the liquid is approximately (In pa.s)

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13000

130

11300

answer is C.

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Detailed Solution

$\begin{array}{l} r = 1 cm; r = 10^{- 2} m \\ ρ = 1 .5 gm / c . c = \frac{1 .5 \times 10^{- 3}}{10^{- 6}} = 1.5 \times 10^{3} kg / m^{3} \\ V_{t} = 0 .25 cm / \sec = 0 .25 \times 10^{- 2} m / \sec; ρ_{m} = 0 \\ V_{t} = \frac{2}{9} \frac{r^{2} g}{η} (ρ_{b} - ρ_{m}) \Rightarrow η = \frac{2}{9} \frac{r^{2} g (ρ_{b} - ρ_{m})}{v_{t}} \\ η = \frac{2}{9} \times 10^{- 4} \times 10 \times \frac{1 .5 \times 10^{3}}{25 \times 10^{- 4}} = 133 Pa . s \end{array}$

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