When the drag force is given by Stokes\' law, F s = 6 rv , and exactly balances

Question

When the drag force is given by Stokes' law, Fs = 6rv, and exactly balances the weight of a falling object the terminal velocity v is given by v=2R2g9(s1), where R is the radius of the sphere, s is its density, and 1 is the density of the fluid and the coefficient of viscosity. Use that equation to find the viscosity of motor oil in which a steel ball of radius 0.8 mm falls with a terminal speed of 4.32 cm/s. The densities of the ball and the oil are 7.86 and 0.88 g/mL, respectively. (answer in mPa s)

Explanation / Answer

Note: Terminal velocity formula mentioned here is wrong, may be typo error, Hereby I provide the exact equation to calculate terminal velocity based on this equation, the following calculation made,

v= {2* R2*g (s1)} / 9

Viscosity of the motor oil () = {2* R2*g (s1)} / 9v

Covert terminal speed in mm/s, so that v= 43.2 mm/s.

                                    = {2* (0.8)2*9.8* (7.86-0.88)} / (9*43.2)

                                    =225 mPa.s

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