Solving the Sphere Volume Problem

The question asks how many smaller solid spherical balls, each with a radius of 33 cm, can be formed by melting down a larger solid spherical ball with a radius of 66 cm. When a solid object is melted and recast into smaller objects, the total volume of the material remains constant.

Understanding the Volume of a Sphere

The volume of a solid sphere is calculated using the formula:

\(V = \frac{4}{3}\pi r^3\)

where \(V\) is the volume and \(r\) is the radius of the sphere.

Calculating Volumes

First, let's calculate the volume of the large solid spherical ball (radius \(R = 66\) cm):

\(V_{large} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (66)^3\)

Next, let's calculate the volume of one small solid spherical ball (radius \(r = 33\) cm):

\(V_{small} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (33)^3\)

Finding the Number of Smaller Spheres

The total volume of the material from the large sphere is conserved when making the smaller spheres. If \(N\) is the number of small spheres that can be made, then the total volume of \(N\) small spheres must equal the volume of the large sphere.

\(N \times V_{small} = V_{large}\)

We can find \(N\) by dividing the volume of the large sphere by the volume of a single small sphere:

\(N = \frac{V_{large}}{V_{small}} = \frac{\frac{4}{3}\pi (66)^3}{\frac{4}{3}\pi (33)^3}\)

The terms \(\frac{4}{3}\pi\) cancel out:

\(N = \frac{(66)^3}{(33)^3}\)

We can rewrite this as:

\(N = \left(\frac{66}{33}\right)^3\)

Simplifying the fraction inside the parenthesis:

\(N = (2)^3\)

Calculating the final value:

\(N = 8\)

Therefore, 8 solid spherical balls each of 33 cm radius can be made out of a solid spherical ball of radius 66 cm.

Summary of Calculations

Number of small spheres = \(\frac{\text{Volume of Large Sphere}}{\text{Volume of Small Sphere}} = \frac{\frac{4}{3}\pi (66)^3}{\frac{4}{3}\pi (33)^3} = \left(\frac{66}{33}\right)^3 = 2^3 = 8\)

Revision Table: Key Concepts

Additional Information: Scaling and Volume

This problem demonstrates how volume scales with linear dimensions. If you double the radius of a sphere, its volume doesn't just double; it increases by a factor of \(2^3 = 8\). In general, if you scale the linear dimensions of a 3D object by a factor \(k\), the volume scales by a factor of \(k^3\). In this case, the radius of the large sphere is twice the radius of the small sphere (\(66/33 = 2\)), so the volume of the large sphere is \(2^3 = 8\) times the volume of the small sphere. This is why 8 smaller spheres can be made from one larger sphere.