Volumes Practice Test 3

Concept:When a solid is melted and recast, its volume remains unchanged. So, equate the volume of the sphere with the volume of the cone.Explanation:Radius of the spherical ball: $r_s = \frac{6}{2} = 3\ \text{cm}$.Volume of the sphere: $V_s = \frac{4}{3}\pi r_s^3 = \frac{4}{3}\pi (3)^3 = 36\pi\ \text{cm}^3$.Radius of the cone base: $r_c = \frac{12}{2} = 6\ \text{cm}$.Let the height of the cone be $h$.Volume of the cone: $V_c = \frac{1}{3}\pi r_c^2 h = \frac{1}{3}\pi (6)^2 h = 12\pi h\ \text{cm}^3$.Since volume is conserved, $V_s = V_c$.$36\pi = 12\pi h$.Cancel $\pi$: $36 = 12h$.$h = \frac{36}{12} = 3\ \text{cm}$.Answer:The height of the cone is $3\ \text{cm}$, so option A is correct.