A steel ball of radius 1.0 mm falls through a tall jar of glycerine between two marks in 30.0 s…

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A steel ball of radius $1.0$ mm falls through a tall jar of glycerine between two marks in $30.0$ s. Another steel ball of radius $1.8$ mm falls between the same marks in the same glycerine. Both reach their steady speeds. Find the ratio of the terminal speed of the second ball to that of the first, and the time that the second ball takes.

Need a reminder? The key ideaShow

Capillary rise: $h = \dfrac{2T\cos\theta}{r\rho g}$ , so $h$ is in proportion to $1/r$ . Stokes' law: $F = 6\pi\eta r v$ . At the terminal speed, $\eta = \dfrac{2r^2(\rho - \sigma)g}{9v_t}$ , and $v_t \propto r^2$ . The Reynolds number is $Re = \dfrac{\sigma v_t(2r)}{\eta}$ . Use SI units.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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