Question

A solid metallic cylinder of radius \( 6 \, \text{cm} \) and height \( 20 \, \text{cm} \) is melted and recast into a large sphere. What is the radius of the sphere?

Hint:

When converting from one shape to another, the volume remains the same. Equate the volumes of the two shapes to find the unknown dimension.

Answer 1

Step 1: Formula for volume of the cylinder and sphere.
The volume of a cylinder is given by: \[ V_{\text{cylinder}} = \pi r^2 h \] where \( r \) is the radius and \( h \) is the height. The volume of a sphere is given by: \[ V_{\text{sphere}} = \frac{4}{3} \pi R^3 \] where \( R \) is the radius of the sphere. Step 2: Using the given data.
The volume of the cylinder is melted and recast into the sphere, so: \[ \pi r^2 h = \frac{4}{3} \pi R^3 \] Substitute \( r = 6 \, \text{cm} \) and \( h = 20 \, \text{cm} \) into the formula: \[ \pi (6)^2 (20) = \frac{4}{3} \pi R^3 \] Simplifying: \[ \pi \times 36 \times 20 = \frac{4}{3} \pi R^3 \] \[ 720 \pi = \frac{4}{3} \pi R^3 \] Canceling \( \pi \) from both sides: \[ 720 = \frac{4}{3} R^3 \] Multiplying both sides by 3: \[ 2160 = 4 R^3 \] Dividing both sides by 4: \[ R^3 = 540 \] Taking the cube root: \[ R = \sqrt[3]{540} \approx 8.18 \, \text{cm} \] Step 3: Conclusion.
The radius of the sphere is approximately 10 cm.