Question

A certain person had a brain that weighed 1.45 kg and contained 6.41 × 1010 cells. assuming that each cell was completely filled with water (density = 1.00 g/ml), calculate the length of one side of such a cell if it were a cube.

Answer 1

Weight of 6.41×1010 number of cells is 1.45 Kg i.e, 1.45×$10^{3}$ gm. So, Weight of 1 (one) number of cells is= $\frac{1.45X10^{3} }{6.41X1010}$ gm = 0.2239 gm.

Assuming that each cell was completely filled with water, from density, Volume of 1 gm water is 1 ml, so volume of 0.2239 gm water is 0.2239 ml. That means, volume of each cell is 0.2239 ml= 0.2239×$10^{-6}$ m³. If each cell is considered as cube, volume of cube=(length of each side of cube)³= 0.2239×$10^{-6}$ m³.

So, length of each side of cube= ∛(0.2239×$10^{-6}$= 6.07×$10^{-3}$ meter.