In order to minimize neutron leakage from a reactor, the ratio of the surface ar
ID: 1512391 • Letter: I
Question
In order to minimize neutron leakage from a reactor, the ratio of the surface area to the volume must be as small as possible. Assume that a sphere of radius a and a cube both have the same volume.
(a) Find the surface-to-volume ratio for the sphere. (Use any variable or symbol stated above as necessary.)
(Asurface, sphere)/ Vsphere = 3/a
(b) Find the surface-to-volume ratio for the cube. (Use any variable or symbol stated above as necessary.)
(Asurface, cube)/ Vcube =
(c) Which of these reactor shapes would have the minimum leakage? the sphere the cube ==> the sphere
I got a and c right but not b. B is not 6/a. B is not 9/(2pi*a). I already tried those.
Explanation / Answer
b) volume of cube, V = (4/3)*pi*a^3
let L is the length of each side.
so, V = L^3
(4/3)*pi*a^3 = L^3
L = a( (4/3)*pi)^(1/3)
surface area of cube = 6*L^2
= 6*(a( (4/3)*pi)^(1/3))^2
= 6*a^2*( (4/3)*pi )^(2/3)
so, surface-to-volume ratio for the cube = 6*a^2/a^3
= 6*a^2*( (4/3)*pi )^(2/3)/ ( (4/3)*pi*a^3)
= 6*( (4/3)*pi )^(2/3)/ ( (4/3)*pi*a)
= 3.722/a