Which box is heavier

Question

There are 2 identical boxes (cubes). First one contains 27 big identical marbles and second one contains 64 small identical marbles. The marbles are made by steel. Supposing that in each box the marbles are next to each other, which box is heavier?

Answer 1

Assuming that both marbles are made of the same material with the same density $\rho$, and also assuming that the same amount of volume is occupied in total by the ($27$ or $64$)marbles in each box.

Let $r$, $R$ be the radius of the smaller and bigger marbles respectively.

Let the cubic box have side $s$ with volume $V = s^3$.

$$\therefore V = 64 \frac{4}{3}\pi r^3 = 27 \frac{4}{3}\pi R^3$$ $$\therefore 64r^3 = 27R^3$$ $$\therefore 4r = 3R$$ $$\therefore \frac{r}{R} = \frac{3}{4}$$ The weight of the box due to the marbles in each case shall be, $$W_1 = 27 W_R , W_2 = 64 W_r$$ $$\therefore \frac{W_1}{W_2} = \frac{27W_R}{64W_r}$$ Since both marble types have the same density, therefore $$\rho_r = \rho_R$$ $$W_r/V_r = W_R/V_R$$ $$W_r/(\frac{4}{3}\pi r^3) = W_R/(\frac{4}{3}\pi R^3)$$ $$\frac{W_r}{W_R} = (\frac{r}{R})^3$$ $$\therefore \frac{W_r}{W_R} = \frac{27}{64}$$ $$\therefore \frac{W_1}{W_2} = 1$$ $$\therefore W_2 = W_1$$

Hence both cubical boxes will weight the same.

Answer 2

Partial answer. If we assume that the marbles are packed in a cubical lattice (as is suggested by the numbers $27 = 3^3$ and $64 = 4^3$), then the smaller marbles have $27/64$ times the volume of the larger ones, but that is exactly compensated for by the fact that there are $64/27$ times as many of them. The marbles in each box thus have the same total weight.

However, if we cannot assume that, and in fact the marbles are as large as possible while still fitting into the box, the problem becomes much more complex. I suspect that the sphere-packing-into-cube problem is not solved for $27$ or $64$ spheres. One would tend to suspect that $64$ spheres could pack into a cube more efficiently than $27$ could (and therefore the $64$-sphere cube would weigh more), but I can't be sure of that.

Answer 3

From 27 to 64, I think the answer is that they are the same. If the number of marbles is small, the most efficient packing is in a square (or cubical) latices. But, when the number of marbles is sufficiently large, dodecahedral structure is more effient (the same pattern your grocer uses to stack oranges).

The dodecahedral structure is less efficient around the walls of the box, and more efficient in the interior of the box. This is why the number of marbles has to pass some minimal condition before they would settle into the more efficient packing. As the number of marbles increases, the theoretical efficiency increases as there are proportionally fewer marbles against the walls of the box.

As the numbers get large you approach some limit of packing efficiency. As I remember 72%