Sphere Packing

What is the length of the shortest pipe, of internal radius $\pu{50 mm}$, that can fully contain $21$ balls of radii $\pu{30 mm}, \pu{31 mm}, \dots, \pu{50 mm}$?
Give your answer in micrometres ($\pu{10^{-6} m}$) rounded to the nearest integer.

First, we need to find the combined length of all 21 balls. This length will be the sum of their diameters, as the balls are assumed to be arranged end-to-end inside the pipe.

The diameter of a sphere is twice its radius. So, the diameter of the ball with radius 30 mm will be $2\times 30 = 60$ mm, for the ball with radius 31 mm the diameter will be $2\times 31 = 62$ mm, and so on up to the ball with radius 50 mm whose diameter will be 100 mm.

The total length of pipes needed is therefore equal to the sum of these diameters from 60 mm to 100 mm in steps of 2 mm.

This forms an arithmetic series with a first term (a) of 60, a common difference (d) of 2, and 21 terms (n).

The sum (S) of an arithmetic series can be calculated using the formula:
S = n/2 [2a + (n – 1)d]

Here:
S = 21/2 [ 2*60 + (21 – 1)*2]
S = 10.5 [120 + 40]
S = 10.5 * 160
S = 1680 mm

Now, we want the answer in micrometers, and there are $10^{3}$ micrometers in 1 millimeter.

So, $1680$ millimeters = $1680 \times 10^{3}$ micrometers = $1680000$ micrometers.

So, the pipeline should be at least $1680000$ micrometers long, rounded to the nearest integer.

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