Understanding the Problem: Making Lead Balls from a Sphere

The question asks how many small lead balls can be made from a single large solid lead sphere. This is a classic problem involving volumes. When a solid material is recast or reshaped into smaller pieces, the total volume of the material remains constant. Therefore, the total volume of all the small lead balls combined must equal the volume of the original large lead sphere.

To solve this, we need to:

1. Calculate the volume of the large lead sphere.
2. Calculate the volume of one small lead ball.
3. Divide the volume of the large sphere by the volume of a small ball to find the number of small balls that can be made.

Calculating Volumes of Spheres

The formula for the volume of a sphere is given by:

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

where \(r\) is the radius of the sphere and \(\pi\) is a mathematical constant (approximately 3.14159).

Volume of the Large Lead Sphere

The diameter of the large lead sphere is given as 42 cm.

The radius (\(R\)) of the large sphere is half of its diameter:

\(R = \frac{\text{Diameter}}{2} = \frac{42 \text{ cm}}{2} = 21 \text{ cm}\)

Now, let's calculate the volume of the large sphere (\(V_{large}\)) using the formula:

\(V_{large} = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (21 \text{ cm})^3\)

Volume of a Small Lead Ball

The diameter of each small lead ball is given as 3 cm.

The radius (\(r\)) of a small lead ball is half of its diameter:

\(r = \frac{\text{Diameter}}{2} = \frac{3 \text{ cm}}{2} = 1.5 \text{ cm}\)

Now, let's calculate the volume of one small lead ball (\(V_{small}\)) using the formula:

\(V_{small} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.5 \text{ cm})^3\)

Finding the Number of Lead Balls

The number of small lead balls that can be made from the large sphere is the ratio of the volume of the large sphere to the volume of one small ball.

Number of balls = \(\frac{V_{large}}{V_{small}}\)

Substitute the volume expressions we found:

Number of balls = \(\frac{\frac{4}{3} \pi (21)^3}{\frac{4}{3} \pi (1.5)^3}\)

Notice that the \(\frac{4}{3} \pi\) terms cancel out from the numerator and the denominator:

Number of balls = \(\frac{(21)^3}{(1.5)^3}\)

We can rewrite this as:

Number of balls = \(\left(\frac{21}{1.5}\right)^3\)

Let's calculate the value inside the parenthesis:

\(\frac{21}{1.5} = \frac{21}{\frac{3}{2}} = 21 \times \frac{2}{3} = \frac{21 \times 2}{3} = \frac{42}{3} = 14\)

So, the number of balls is:

Number of balls = \((14)^3\)

Now, calculate \(14^3\):

\(14^3 = 14 \times 14 \times 14\)

\(14 \times 14 = 196\)

\(196 \times 14 = 196 \times (10 + 4) = (196 \times 10) + (196 \times 4)\)

\(196 \times 10 = 1960\)

\(196 \times 4\):

\(196 \times 4 = 400 + 360 + 24 = 784\)

So, \(14^3 = 1960 + 784 = 2744\)

Therefore, 2744 lead balls, each 3 cm in diameter, can be made from a solid lead sphere of diameter 42 cm.

Summary of Calculations

Number of balls = \(\frac{\text{Volume of large sphere}}{\text{Volume of small ball}} = \frac{(21)^3}{(1.5)^3} = \left(\frac{21}{1.5}\right)^3 = (14)^3 = 2744\)

Revision Table: Volume of Sphere Concepts

Additional Information: Scaling and Volumes

This problem demonstrates an important concept about how volume scales with linear dimensions. When you change the size of an object, its volume changes according to the cube of the scaling factor of its linear dimensions (like radius or diameter).

In this case, the ratio of the diameters is \(\frac{42 \text{ cm}}{3 \text{ cm}} = 14\). This means the large sphere's diameter is 14 times the small ball's diameter. Since radius is directly proportional to diameter, the ratio of radii is also 14 (\(\frac{21 \text{ cm}}{1.5 \text{ cm}} = 14\)).

The ratio of the volumes is the cube of the ratio of the linear dimensions (radii or diameters):

\(\frac{V_{large}}{V_{small}} = \left(\frac{R}{r}\right)^3 = \left(\frac{21}{1.5}\right)^3 = (14)^3\)

This relationship holds true for any two similar 3D shapes. If one shape has linear dimensions \(k\) times larger than another similar shape, its volume will be \(k^3\) times larger.