Mathematics

Mathematics

Problem

A cylindrical bucket has an internal volume of 1.0×10−2 m31.0 \times 10^{-2}\,\text{m}^{3}. It is filled with dry sand whose grains can be approximated as spheres of diameter 5.0×10−4 m5.0 \times 10^{-4}\,\text{m}. Because of void spaces, sand grains occupy only 60%60\% of the bucket’s volume. The order of magnitude of the number of grains that can fit into the bucket is a) 10710^{7} b) 10810^{8} c) 10610^{6} d) 10910^{9}
Intuitive explanation

The key insight is that you're not counting individual grains directly—you're using the packing efficiency to find how much of the bucket is actually occupied by sand material, then dividing by the volume of one grain. Calculate the volume of a single spherical grain using V=43πr3V = \frac{4}{3}\pi r^3, then use the fact that sand grains fill 60% of the total bucket volume to find how many grains fit. This combines geometric volume formulas with the practical reality that randomly packed spheres leave gaps.

Intuitive explanation

The key insight is that you're not counting individual grains directly—you're using the packing efficiency to find how much of the bucket is actually occupied by sand material, then dividing by the volume of one grain. Calculate the volume of a single spherical grain using V=43πr3V = \frac{4}{3}\pi r^3, then use the fact that sand grains fill 60% of the total bucket volume to find how many grains fit. This combines geometric volume formulas with the practical reality that randomly packed spheres leave gaps.

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