If an air bubble of diameter 2 text{ mm} rise | vedclass

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(B) The terminal velocity $v$ of an air bubble rising through a liquid is given by Stokes' Law: $v = \frac{2 r^2 g (\rho_l - \rho_g)}{9 \eta}$.Since t

Vedclass · Accepted Answer

$8.88$

Vedclass · Answer

(B) The terminal velocity $v$ of an air bubble rising through a liquid is given by Stokes' Law: $v = \frac{2 r^2 g (ho_l - ho_g)}{9 \eta}$.Since the density of air $ho_g$ is negligible compared to the liquid density $ho_l$, we use $ho_l = 2000 	ext{ kg/m}^3$.Given: diameter $d = 2 	ext{ mm} \implies r = 1 	ext{ mm} = 10^{-3} 	ext{ m}$, $v = 0.5 	ext{ cm/s} = 0.5 	imes 10^{-2} 	ext{ m/s}$, and $g = 10 	ext{ m/s}^2$.Rearranging for viscosity $\eta$: $\eta = \frac{2 r^2 g ho_l}{9 v}$.Substituting the values: $\eta = \frac{2 	imes (10^{-3})^2 	imes 10 	imes 2000}{9 	imes 0.5 	imes 10^{-2}} = \frac{2 	imes 10^{-6} 	imes 20000}{4.5 	imes 10^{-2}} = \frac{0.04}{0.045} \approx 0.888 	ext{ Pa.s}$.Since $1 	ext{ Pa.s} = 10 	ext{ Poise}$, $\eta = 0.888 	imes 10 = 8.88 	ext{ Poise}$.

If an air bubble of diameter $2 \text{ mm}$ rises steadily through a liquid of density $2000 \text{ kg/m}^3$ at a rate of $0.5 \text{ cm/s}$, then the coefficient of viscosity of the liquid is . . . . . . $\text{Poise}$. (Take $g = 10 \text{ m/s}^2$)

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