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(C) The terminal velocity $v_T$ of an air bubble rising in a liquid is given by the formula:$v_T = \frac{2}{9} \frac{r^2 g (\rho_l - \rho_a)}{\eta}$Gi

Vedclass · Accepted Answer

$1.66 	imes 10^3\, 	ext{poise}$

Vedclass · Answer

(C) The terminal velocity $v_T$ of an air bubble rising in a liquid is given by the formula:$v_T = \frac{2}{9} \frac{r^2 g (ho_l - ho_a)}{\eta}$Given that the density of air $ho_a$ is negligible,the formula becomes:$v_T = \frac{2}{9} \frac{r^2 g ho_l}{\eta}$Given values:Radius $r = 1\, cm$Terminal velocity $v_T = 2.00\, mm/sec = 0.2\, cm/sec$Density of liquid $ho_l = 1.5\, g/cm^3$Acceleration due to gravity $g = 1000\, cm/sec^2$Substituting these values into the equation:$0.2 = \frac{2}{9} 	imes (1)^2 	imes \frac{1000 	imes 1.5}{\eta}$$0.2 = \frac{2 	imes 1500}{9 	imes \eta}$$0.2 = \frac{3000}{9 \eta}$$0.2 = \frac{1000}{3 \eta}$$\eta = \frac{1000}{3 	imes 0.2} = \frac{1000}{0.6} = \frac{10000}{6} \approx 1666.67\, 	ext{poise}$$\eta \approx 1.66 	imes 10^3\, 	ext{poise}$

An air bubble of $1\, cm$ radius is rising at a steady rate of $2.00\, mm/sec$ through a liquid of density $1.5\, g/cm^3$. Neglect the density of air. If $g = 1000\, cm/sec^2$,then the coefficient of viscosity of the liquid is:

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