Understanding the Problem: Lead Sphere Conversion

The question asks us to determine how many smaller solid lead balls of a specific diameter can be created from a larger solid lead sphere. This type of problem involves understanding the concept of volume and how volume is conserved when a material is reshaped or divided.

When a solid sphere is melted down and recast into smaller spheres, the total volume of the material remains the same. Therefore, the sum of the volumes of all the smaller spheres must be equal to the volume of the original large sphere.

Calculating Sphere Dimensions: Radius from Diameter

The volume of a sphere depends on its radius. The radius is always half of the diameter.

• Diameter of the large sphere = 14 cm
• Radius of the large sphere (R) = Diameter / 2 = 14 cm / 2 = 7 cm

• Diameter of each small ball = 2 cm
• Radius of each small ball (r) = Diameter / 2 = 2 cm / 2 = 1 cm

Sphere Volume Formula

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

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

Where:

• \(V\) is the volume of the sphere
• \(\pi\) is a mathematical constant (approximately 3.14159)
• \(r\) is the radius of the sphere

Calculating Volumes

Now, we will calculate the volume of the large sphere and the volume of one small lead ball using the formula:

Volume of the large sphere:

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

\(V_{\text{large}} = \frac{4}{3}\pi (7 \text{ cm})^3\)

\(V_{\text{large}} = \frac{4}{3}\pi (7 \times 7 \times 7 \text{ cm}^3)\)

\(V_{\text{large}} = \frac{4}{3}\pi (343 \text{ cm}^3)\)

Volume of one small ball:

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

\(V_{\text{small}} = \frac{4}{3}\pi (1 \text{ cm})^3\)

\(V_{\text{small}} = \frac{4}{3}\pi (1 \times 1 \times 1 \text{ cm}^3)\)

\(V_{\text{small}} = \frac{4}{3}\pi (1 \text{ cm}^3)\)

Finding the Number of Small Balls

Since the total volume of lead is conserved, the number of small balls that can be made is equal to the total volume of the large sphere divided by the volume of a single small ball.

Let \(n\) be the number of small balls.

\(n = \frac{\text{Volume of large sphere}}{\text{Volume of one small ball}}\)

\(n = \frac{\frac{4}{3}\pi (343 \text{ cm}^3)}{\frac{4}{3}\pi (1 \text{ cm}^3)}\)

Notice that the term \(\frac{4}{3}\pi\) appears in both the numerator and the denominator. This term cancels out, simplifying the calculation significantly.

\(n = \frac{343 \text{ cm}^3}{1 \text{ cm}^3}\)

\(n = 343\)

Therefore, 343 solid lead balls each 2 cm in diameter can be made from a solid sphere made up of lead of diameter 14 cm.

Revision Table: Sphere Volume Calculation

Additional Information: Volume Conservation in Geometry

The principle of volume conservation is fundamental in many geometry and physics problems. When a substance changes shape but its mass and density remain constant (like melting lead and recasting it), its volume also remains constant. This principle is used in various scenarios, such as:

• Calculating the height of water in a cylindrical container when a sphere is submerged.
• Finding the number of smaller cones or cylinders that can be formed from melting a larger solid shape.
• Determining the change in water level when ice (which displaces water based on its mass) melts into liquid water.

In this specific problem, the ratio of the volumes is simply the cube of the ratio of the radii (or diameters), because the \(\frac{4}{3}\pi\) term cancels out:

\(n = \frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3} = \frac{R^3}{r^3} = \left(\frac{R}{r}\right)^3 = \left(\frac{D/2}{d/2}\right)^3 = \left(\frac{D}{d}\right)^3\)

Using the diameters directly: \(n = \left(\frac{14 \text{ cm}}{2 \text{ cm}}\right)^3 = (7)^3 = 7 \times 7 \times 7 = 343\).

This shortcut works specifically because we are dealing with similar shapes (spheres) and dividing the volume of the larger shape by the volume of the smaller shape to find the number of smaller shapes.