|
Question 1203304: a solid metal sphere is melted and recast into a hollow spherical shell whose outer radius is 20cm. if the radius of the hollow interior of the shell is equal to the radius of the original sphere, what is the radius of the original sphere?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the smaller sphere is solid.
the volume of that sphere is 4/3 * pi * r^3.
the larger sphere has a hollow center.
that hollow center has the same volume as the original sphere.
that means that the volume of the filled shell of the larger sphere is the same volume of the original sphere.
if you let the total volume of the larger spghere equal to V2 and you let the volume of the original sphere equal to V1, then you get:
V2 = 2 * V1.
V1 is the volume of the hollow center of the larger sphere.
V1 is also the volume of the filled shell part of the larger sphere.
you know that V2 = 4/3 * pi * 20^3.
that's the volume of the hollow center plus the filled shell, i.e. it is the total volume of the larger sphere.
your formula is therefore V2 = 2 * V1.
this becomes:
4/3 * pi * 20^3 = 2 * 4/3 * pi * r^3.
r is the radius of V1.
if you divide both sides of the equation by (2 * 4/3 * pi), you get:
20^3 / 2 = r^3
since 20^3 = 8000, then you get:
r^3 = 4000
solve for r to get:
r = 4000 ^ (1/3) = 15.87401052.
that's your solution.
with that radius, the volume of the hollow center is equal to 4/3 * pi * 15.87401052^3 = 16755.16082.
since you know that V2 = 2 * V1, then V2 = 33510.62164.
that's the same as r/3 * pi * 20^3 = 33510.32164, so you're good.
your solution is, best that i can determine, equal to 15.87401052.
|
|
|
| |
