Understanding the Problem: Sphere to Cone Recasting

The problem describes a situation where a solid metallic sphere is melted down and then reshaped into a cone. The key principle here is that when a solid object is melted and reformed into another shape, its volume remains constant. We are given the radius of the sphere and the height of the cone, and we need to find the base radius of the cone.

Volume Conservation Principle

When a solid is melted and recast, the volume of the material does not change. Therefore, the volume of the original sphere is equal to the volume of the new cone.

\( \text{Volume of Sphere} = \text{Volume of Cone} \)

Formulas for Volume

We need the formulas for the volume of a sphere and a cone:

• Volume of a sphere: \( V_{sphere} = \frac{4}{3}\pi r^3 \) where \( r \) is the radius of the sphere.
• Volume of a cone: \( V_{cone} = \frac{1}{3}\pi R^2 h \) where \( R \) is the base radius of the cone and \( h \) is the height of the cone.

Given Information

From the question, we have:

• Radius of the sphere, \( r = 8 \) cm
• Height of the cone, \( h = 8 \) cm

We need to find the base radius of the cone, \( R \).

Step-by-Step Calculation of Cone Radius

Using the principle of volume conservation, we set the volume of the sphere equal to the volume of the cone:

\( V_{sphere} = V_{cone} \)

\( \frac{4}{3}\pi r^3 = \frac{1}{3}\pi R^2 h \)

Substitute the given values for \( r \) and \( h \):

\( \frac{4}{3}\pi (8)^3 = \frac{1}{3}\pi R^2 (8) \)

We can cancel \(\frac{1}{3}\pi\) from both sides of the equation:

\( 4 \times (8)^3 = R^2 \times 8 \)

Calculate \(8^3\):

\( 4 \times 512 = 8 R^2 \)

Multiply 4 by 512:

\( 2048 = 8 R^2 \)

Now, isolate \( R^2 \) by dividing both sides by 8:

\( R^2 = \frac{2048}{8} \)

\( R^2 = 256 \)

To find \( R \), take the square root of 256:

\( R = \sqrt{256} \)

\( R = 16 \)

So, the base radius of the cone is 16 cm.

Summary of Calculation

The base radius of the cone is 16 cm.

Revision Table: Key Concepts in Recasting Problems

Additional Information: Mensuration of 3D Shapes

This problem falls under the category of Mensuration, specifically dealing with the volumes of three-dimensional shapes. Understanding the formulas for common solids is crucial for solving such problems. Here are some other important shapes and their volume formulas:

• Cylinder: \(V = \pi r^2 h\) (where \(r\) is base radius, \(h\) is height)
• Cube: \(V = a^3\) (where \(a\) is side length)
• Cuboid: \(V = l \times w \times h\) (where \(l\), \(w\), \(h\) are length, width, height)
• Pyramid: \(V = \frac{1}{3} \times \text{Base Area} \times h\) (where \(h\) is height)

Problems involving melting and recasting, or transferring liquids between containers, often rely on the principle of volume conservation. Always identify the initial and final shapes and use their respective volume formulas to set up an equation.