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The size of spherical object with the weight of 38 pounds can be calculated if you know the density.

The question states "stone".

I prefer to perform such calculations in SI units (kilograms and meters) because it its SO much easier but just here I'll try using Avoirdupois and inches.

The density of "stone" is not precise and as "geology.about.com" states:

Rock density is very sensitive to the minerals that compose a particular rock type. Sedimentary rocks (and granite), which are rich in quartz and feldspar, tend to be less dense than volcanic rocks. And if you know your igneous petrology, you'll see that the more mafic a rock is, the greater its density.

We will assume a density of 1,8 ounces per cubic inch (about 3 grams per cubic centimeter)

38 pounds is 608 ounces

608 / 1,8 is about 338 (that is the volume in cubic inches of the stone)

V=4/3 pi r3 is the formula for the volume of a sphere

338 = 4/3 * 3.1416 * r3

80.6 = r3

4,32 = r

The diameter of the 38 pound spherical stone is approximately 8,6 inches

Denser stone would give a smaller diameter, less dens stone a larger diameter.

Q: What is the approximate diameter of these 38-pound stones?

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It is about 12 feet in diameter. Hope this helps!