4. A ten-liter bucket was filled to the brim with currants. Gavrila immediately said that there were 10 kg of currants in the bucket. Glafira thought about it and estimated the weight of the berries in the bucket more accurately. How can this be done if the density of the currant can be approximately considered equal to the density of water?
Solution. In approximate calculations, the sizes of the berries can be considered the same and much smaller than the size of the bucket. If the berries are laid out in one layer, then in the densest packing, each berry will have 6 neighbors: the centers of the berries will be at the vertices of equilateral triangles with sides equal to the diameter of the berries. When pouring the next layer, the berries will be located in the depressions between the berries of the previous layer. In such a packing, each berry will have 12 neighbors, and the centers of the berries will be at the vertices of regular tetrahedrons.

To calculate the fraction of the volume occupied by the berries, one can mentally "cut out" a parallelepiped from the lattice formed by the centers of the berries with edges of length $2 N R$, where $R$ is the radius of the berries, and $N$ is the number of berries that fit along one side, and in two opposite vertices of the edge, they meet at an angle of $60^{\circ}$. The volume of such a parallelepiped is $4 \sqrt{2} N^{3} R^{3}$, and it contains approximately $N^{3}$ berries, each with a volume of $\frac{4}{3} \pi R^{3}$ (berries near the walls are not fully placed, but their number is of the order of $N^{2}$, so this can be neglected for large $N$). Therefore, the ratio of the volume of the berries to the volume of the bucket is approximately $\frac{\pi \sqrt{2}}{6} \approx 0.74$, so in a 10-liter bucket, with the densest packing, there are about 7.4 kg of berries. In reality, there are somewhat fewer due to the lack of density.

The ratio of the volume of the berries to the volume of the container can also be calculated differently. For example, consider a rectangular parallelepiped in which the berries are laid out in layers, the distance between which is equal to the height of the tetrahedron mentioned above $\frac{2 \sqrt{2} R}{\sqrt{3}}$. In each layer, the distance between rows is equal to the height of the equilateral triangle $R \sqrt{3}$. Thus, $N^{3}$ berries will occupy a parallelepiped with a length of $2 R N$, a width of $\sqrt{3} R N$, and a height of $\frac{2 \sqrt{2} R N}{\sqrt{3}}$. The ratio of volumes is $\frac{N^{3} \cdot \frac{4}{3} \pi R^{3}}{2 R N \cdot \sqrt{3} R N \cdot \frac{2 \sqrt{2} R N}{\sqrt{3}}}=\frac{\pi}{3 \sqrt{2}}=\frac{\pi \sqrt{2}}{6}$.

Answer: about 7 kg.

Criteria: 20 points - correct (not necessarily the same as above) solution and the correct answer (from 7 to 7.4 kg); 15 points - generally ideologically correct solution, with errors that did not affect the answer; 10 points - ideologically correct solution with significant errors; also evaluated are solutions in which the packing density coefficient $\frac{\pi \sqrt{2}}{6}$ (or 74%) is used without proof; 5 points - the packing in the layer is taken into account, but the packing between layers is not; **0** points - everything else.