$5\mathrm{.}4$ Time Required for Sedimentation by Gravity A certain reagent is added to a suspension of cells $4\backslash$mu m in diameter. These cells have a density of $1\mathrm{.}08$ g$/$cm$3,$ and they are suspended in liquid with a density of $1\mathrm{.}00$ g$/$cm$3$ and viscosity of $1\mathrm{.}0$ cp$.$ This reagent causes about half of the cells to form fairly solid aggregates, all of which are $90\backslash$mu m in diameter and have density midway between that of the liquid and the cells. How much time is required for all the aggregates to sediment to within $1$ cm of the bottom of a vessel filled with suspension that is $0\mathrm{.}5$ m high? Approximately what fraction of the single cells would have sedimented to this depth in the same amount of time? How much time is required for all the single cells to sediment to within $1$ cm of the bottom of the vessel?

Answer should be

Time for aggregates $=46\mathrm{.}2$ min

Fraction $=0\mathrm{.}02+0\mathrm{.}0039=0\mathrm{.}024$

Time for all cells $=195$ h

Hint:

Remember: Settling time $=$ Settling distance $/$ Settling velocity

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For the single cell sedimentation $1/50$ of the cells are already within $1$ cm of the bottom $=>$ you have to add the fraction which additional settles.