To solve this problem, we will use the concept of volume conservation since the material of the cylinder is melted to form a cone.

Step 1: Volume of the Cylinder

The formula for the volume of a cylinder is:

$V_{\text{cylinder}} = \pi r^2 h$

Here:

• Base radius of the cylinder $r = 8 \, \text{cm}$
• Height of the cylinder $h = 2 \, \text{cm}$

$V_{\text{cylinder}} = \pi (8)^2 (2) = \pi \cdot 64 \cdot 2 = 128 \pi \, \text{cm}^3$

Step 2: Volume of the Cone

The formula for the volume of a cone is:

$V_{\text{cone}} = \frac{1}{3} \pi R^2 H$

Here:

• Height of the cone $H = 6 \, \text{cm}$
• Radius of the cone $R$ (to be determined)

Equating the volumes of the cylinder and cone:

$128 \pi = \frac{1}{3} \pi R^2 (6)$

Cancel $\pi$ from both sides:

$128 = 2 R^2$

Step 3: Solve for $R$

$R^2 = \frac{128}{2} = 64$

$R = \sqrt{64} = 8 \, \text{cm}$

Final Answer:

The radius of the cone is 8 cm.

Would you like further details or explanations? Here are related questions for deeper understanding:

1. How does volume conservation help in solving such problems?
2. What are the key differences between the formulas for the volume of a cylinder and a cone?
3. How would the result change if the height of the cone were doubled?
4. Can this method be extended to irregular shapes?
5. Why is the factor $\frac{1}{3}$ present in the cone volume formula?

Tip: Always ensure units are consistent when working with volume problems!