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(B) As the bubble is rising steadily,the net force acting on it is zero.Since the bubble is rising,the buoyant force $(B)$ acts upwards,while the visc

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) As the bubble is rising steadily,the net force acting on it is zero.Since the bubble is rising,the buoyant force $(B)$ acts upwards,while the viscous drag $(F)$ and the weight $(mg)$ act downwards. Given the density of air is negligible,$mg \approx 0$.Therefore,$B = F$.Using Stokes' Law for viscous drag,$F = 6 \pi \eta R v$,and the buoyant force $B = \frac{4}{3} \pi R^3 ho g$.Equating them: $\frac{4}{3} \pi R^3 ho g = 6 \pi \eta R v$.Solving for $\eta$: $\eta = \frac{2 R^2 ho g}{9 v}$.Given: $R = 1\,mm = 10^{-3}\,m$,$ho = 1750\,kg\,m^{-3}$,$g = 9.8\,m\,s^{-2}$ (or $10\,m\,s^{-2}$),$v = 0.35\,cm\,s^{-1} = 0.35 	imes 10^{-2}\,m\,s^{-1}$.Substituting values: $\eta = \frac{2 	imes (10^{-3})^2 	imes 1750 	imes 10}{9 	imes 0.35 	imes 10^{-2}} = \frac{2 	imes 10^{-6} 	imes 17500}{3.15 	imes 10^{-2}} = \frac{0.035}{0.0315} \approx 1.11\,Pa\,s$.Since $1\,Pa\,s = 10\,poise$,$\eta = 1.11 	imes 10 = 11.1\,poise$.The nearest integer is $11$.

The diameter of an air bubble,which was initially $2\,mm$,rises steadily through a solution of density $1750\,kg\,m^{-3}$ at the rate of $0.35\,cm\,s^{-1}$. The coefficient of viscosity of the solution is (in poise,nearest integer). (The density of air is negligible).

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