Question

A cylinder of radius \( 12 \, \text{cm} \) contains water up to the height \( 20 \, \text{cm} \). A spherical iron ball is dropped into the cylinder, and thus the water level is raised by \( 6.75 \, \text{cm} \). What is the radius of the iron ball?

Hint:

To solve such problems, equate the volume of the displaced water to the volume of the submerged object.

Answer 1

Step 1: The volume of water displaced is equal to the volume of the iron ball: \[ V_{\text{displaced}} = V_{\text{sphere}}. \] Step 2: Volume of water displaced: \[ V_{\text{displaced}} = \pi r^2 h = \pi (12)^2 (6.75) = 972 \pi \, \text{cm}^3. \] Step 3: Volume of the sphere: \[ V_{\text{sphere}} = \frac{4}{3} \pi r^3. \] Equating the volumes: \[ \frac{4}{3} \pi r^3 = 972 \pi. \] Step 4: Simplify: \[ r^3 = \frac{972 \times 3}{4} = 729 \quad \Rightarrow \quad r = \sqrt[3]{729} = 9 \, \text{cm}. \]