Question

A solid sphere of radius 12 inches is melted and cast into a right circular cone whose base diameter is \( \sqrt{2} \) times its slant height. If the radius of the sphere and the cone are the same, how many such cones can be made and how much material is left out?

• A. 4 and 1 cubic inch
• B. 3 and 12 cubic inches
• C. 4 and 0 cubic inch
• D. 3 and 6 cubic inches

Hint:

Use geometry formulas with care and equate volumes for melt-cast problems. Watch for radius/slant height relations.

Answer 1

The Correct Option is C

Volume of sphere: \[ V_s = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (12)^3 = \frac{4}{3} \pi \cdot 1728 = 2304 \pi \] Let cone have radius \( r = 12 \), slant height \( l \), and diameter = \( \sqrt{2} \cdot l \Rightarrow 2r = \sqrt{2}l \Rightarrow l = \frac{24}{\sqrt{2}} = 12\sqrt{2} \) Use Pythagoras: \[ h^2 + r^2 = l^2 \Rightarrow h^2 + 144 = 288 \Rightarrow h^2 = 144 \Rightarrow h = 12 \] So volume of cone: \[ V_c = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (12)^2 \cdot 12 = \frac{1}{3} \pi \cdot 144 \cdot 12 = 576 \pi \] Number of cones = \( \frac{2304\pi}{576\pi} = \boxed{4} \), and leftover = 0