First slide

Viscosity

Question

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

Moderate

A

13,000
B

11,300
C

130
D

13

Solution

r = 1cm; r = 10^-2m; V_t = 0.25cm/sec = 0.25 x 10^-2m/sec; ρ_m = 0
$\large \rho=1.5gm/c.c=\frac {1.5\times 10^{-3}} {10^{-6}}$
when the bubble attain terminal velocity, resultant force on it must be zero.
$\large \therefore$ Viscous drag - Buoyant force = 0
$\large \Rightarrow 6\pi\eta rV_t-\frac 43\pi r^3\rho g=0$
$\large \Rightarrow \eta =\frac 29\frac {\rho g r^2}{V_t}$
$\large \Rightarrow \eta =\frac 29\times 10^{-4}\times 10\times \frac {1.5\times 10^3}{25\times 10^{-4}}=130\;Pa-s$

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