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(D) The gas bubble is rising through the liquid of density $\rho = 1.75 \ g/cm^3$ with a constant terminal velocity $v_T = 0.35 \ cm/s$.Since the dens

Vedclass · Accepted Answer

$1089$

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(D) The gas bubble is rising through the liquid of density $ho = 1.75 \ g/cm^3$ with a constant terminal velocity $v_T = 0.35 \ cm/s$.Since the density of the gas is negligible,the forces acting on the bubble are the buoyancy force $F_b$ (upward) and the viscous drag force $F_V$ (downward).At terminal velocity,the net force is zero,so $F_V = F_b$.Using Stokes' Law,the viscous force is $F_V = 6 \pi \eta r v_T$,where $\eta$ is the coefficient of viscosity and $r$ is the radius of the bubble.The buoyancy force is $F_b = V ho g = \frac{4}{3} \pi r^3 ho g$.Equating the two: $6 \pi \eta r v_T = \frac{4}{3} \pi r^3 ho g$.Solving for $\eta$: $\eta = \frac{2}{9} \frac{r^2 ho g}{v_T}$.Given: $r = 1 \ cm = 10^{-2} \ m$,$ho = 1.75 \ g/cm^3 = 1750 \ kg/m^3$,$v_T = 0.35 \ cm/s = 3.5 	imes 10^{-3} \ m/s$,$g = 9.8 \ m/s^2$.Using $CGS$ units for convenience: $r = 1 \ cm$,$ho = 1.75 \ g/cm^3$,$v_T = 0.35 \ cm/s$,$g = 980 \ cm/s^2$.$\eta = \frac{2}{9} 	imes \frac{(1)^2 	imes 1.75 	imes 980}{0.35} = \frac{2}{9} 	imes \frac{1715}{0.35} = \frac{2}{9} 	imes 4900 \approx 1088.8 \ 	ext{poise} \approx 1089 \ 	ext{poise}$.

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

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