Calculating Small Sphere Diameter from Large Sphere Volume

The problem involves finding the diameter of small spheres formed by recasting a larger sphere, where the total volume remains constant.

Volume Formulas

The volume of a sphere is given by the formula \(V = \frac{4}{3}\pi r^3\), where \(r\) is the radius.

Relating Volumes

Let \(R\) be the radius of the large sphere and \(r\) be the radius of each small sphere. Let \(n\) be the number of small spheres.

• Radius of the large sphere, \(R = 40\) cm.
• Number of small spheres, \(n = 512\).
• Volume of the large sphere: \(V_{large} = \frac{4}{3}\pi R^3\).
• Volume of one small sphere: \(V_{small} = \frac{4}{3}\pi r^3\).
• Total volume of small spheres: \(V_{total\_small} = n \times V_{small} = 512 \times \frac{4}{3}\pi r^3\).

Equating Volumes and Solving for Radius

The problem states that the total volume of the small spheres equals the volume of the large sphere:

\( V_{total\_small} = V_{large} \)

\( 512 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \)

Cancel out the common terms (\(\frac{4}{3}\pi\)):

\( 512 \times r^3 = R^3 \)

Substitute the value of \(R = 40\) cm:

\( 512 \times r^3 = (40)^3 \)

Solve for \(r^3\):

\( r^3 = \frac{(40)^3}{512} \)

Since \(512 = 8^3\), we can write:

\( r^3 = \frac{(40)^3}{8^3} = \left(\frac{40}{8}\right)^3 = (5)^3 \)

Therefore, the radius of each small sphere is \(r = 5\) cm.

Calculating Diameter

The diameter (\(d\)) of a sphere is twice its radius (\(d = 2r\)).

\( d = 2 \times r = 2 \times 5 \text{ cm} = 10 \text{ cm} \)

The diameter of each small sphere is 10 cm.