You are here:

Advanced Math/Diameter, Weight

Advertisement


Question
A solid metal ball, 6 inches in diameter, weight 32 lbs., along with another solid ball of the same metal, 4 inches in diameter, (weight not given), are melted together to form another ball.  

Determine diameter and weight of the new ball.

Thanks.

Answer
Radius of first ball = 6/2 = 3 inches
volume =   (4/3)πr₁³   
    = (4/3)π(3³)   
    ≅ 113.097 in³

Density of metal:
32 lbs/113.097 in³ = (32/113.097) lb/in³

Radius of smaller ball = 4/2 = 2 inches
volume =   (4/3)πr₂³
    = (4/3)π(2³)   
    ≅ 33.510 in³

33.510 in³ ×  (32/113.097) lb/in³ ≅ 9.481 lbs
The second ball weighs 9.481 lbs.

Third ball weighs 32 + 9.481 = 41.481 lbs.
volume = 41.481 lbs × 113.097 in³/(32 lbs) = 146.608 in³
(4/3)πr₃³ = 146.609 in³
r₃³ = 146.609×3/(4π)
r₃ = ∛(146.609×3/(4π)) ≅  3.271 inches
Diameter of third ball = 6.542 inches

Advanced Math

All Answers


Answers by Expert:


Ask Experts

Volunteer


Janet Yang

Expertise

I can answer questions in Algebra, Basic Math, Calculus, Differential Equations, Geometry, Number Theory, and Word Problems. I would not feel comfortable answering questions in Probability and Statistics or Topology because I have not studied these in depth.

Experience

I tutor students (fifth through twelfth grades) and am a Top Contributor on Yahoo!Answers with over 24,000 math solutions.

Publications
Co-author of An Outline of Scientific Writing: For Researchers With English as a Foreign Language

Education/Credentials
I have a Bachelor's degree in Applied Mathematics from the University of California at Berkeley.

Past/Present Clients
George White Elementary School. Homework Help program at the Ridgewood Public Library, Ridgewood, NJ. Individual students.

©2016 About.com. All rights reserved.