An air bubble of diameter 6,mm rises steadily | vedclass

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(B) Since the air bubble is moving at a constant terminal velocity,the net force acting on it is zero.The upward buoyant force $B$ is balanced by the

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$10$

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(B) Since the air bubble is moving at a constant terminal velocity,the net force acting on it is zero.The upward buoyant force $B$ is balanced by the downward viscous drag force $F_v$.$B = F_v$Using Archimedes' principle for buoyancy and Stokes' law for viscous drag:$\frac{4}{3} \pi R^3 ho g = 6 \pi \eta R v$Where $R$ is the radius of the bubble,$ho$ is the density of the solution,$\eta$ is the coefficient of viscosity,and $v$ is the terminal velocity.Given:Diameter $d = 6\,mm \implies R = 3\,mm = 3 	imes 10^{-3}\,m$Density $ho = 1750\,kg/m^3$Velocity $v = 0.35\,cm/s = 0.35 	imes 10^{-2}\,m/s$Acceleration due to gravity $g = 10\,m/s^2$Rearranging for $\eta$:$\eta = \frac{2 R^2 ho g}{9 v}$Substituting the values:$\eta = \frac{2 	imes (3 	imes 10^{-3})^2 	imes 1750 	imes 10}{9 	imes 0.35 	imes 10^{-2}}$$\eta = \frac{2 	imes 9 	imes 10^{-6} 	imes 17500}{9 	imes 0.35 	imes 10^{-2}}$$\eta = \frac{2 	imes 10^{-6} 	imes 17500}{0.35 	imes 10^{-2}}$$\eta = \frac{0.035}{0.0035} = 10\,Pa\cdot s$

An air bubble of diameter $6\,mm$ rises steadily through a solution of density $1750\,kg/m^3$ at the rate of $0.35\,cm/s$. The coefficient of viscosity of the solution (neglect density of air) is $..........\,Pa\cdot s$ (given,$g = 10\,m/s^2$).

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