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(A) According to Stokes' Law,the viscous drag force $F$ acting on a spherical object of radius $r$ moving with terminal velocity $v$ in a fluid of vis

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$8.48 	imes 10^{-3}\,N$

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(A) According to Stokes' Law,the viscous drag force $F$ acting on a spherical object of radius $r$ moving with terminal velocity $v$ in a fluid of viscosity $\eta$ is given by:$F = 6 \pi \eta r v$Given values:Radius $r = 5\,mm = 5 	imes 10^{-3}\,m$Velocity $v = 10\,cm\,s^{-1} = 10 	imes 10^{-2}\,m\,s^{-1} = 0.1\,m\,s^{-1}$Viscosity $\eta = 0.9\,kg\,m^{-1}s^{-1}$Substituting the values into the formula:$F = 6 	imes 3.14 	imes 0.9 	imes (5 	imes 10^{-3}) 	imes (0.1)$$F = 6 	imes 3.14 	imes 0.9 	imes 5 	imes 10^{-4}$$F = 84.78 	imes 10^{-4}\,N$$F = 8.478 	imes 10^{-3}\,N \approx 8.48 	imes 10^{-3}\,N$

The terminal velocity of a copper ball of radius $5\,mm$ falling through a tank of oil at room temperature is $10\,cm\,s^{-1}$. If the viscosity of oil at room temperature is $0.9\,kg\,m^{-1}s^{-1}$,the viscous drag force is:

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