銅線電阻的問題?

Let r1 and r2 be the radius of the cross-sectional area of A and B respectively
since mass = density x volume
Ma/Mb = p(4/3)(pi)(r1)^3/[p(4/3)(pi)(r2)^3]
where Ma and Mb are the masses of copper wire A and B respectively, and p is the density of copper
i.e. Ma/Mb = (r1/r2)^3 = 4/9
(r1/r2)= (4/9)^(1/3) ------------------------------ (1)

Ratio of resistances Ra/Rb = [(ro)L/Aa]/[(ro)L/Ab]
where ro is the resistivity of copper
Aa and Ab are the cross-sectional area of A and B
L are the lengths of A and B

Hence, Ra/Rb = (Ab/Aa) = (pi.(r2)^2)/(pi.(r1)^2) = (r2/r1)^2
using equation (1)
Ra/Rb = (9/4)^(2/3) = 1.717



2011-03-11 13:59:20 補充：
Sorry. .I made a wrong calculation.
Since mass is proportional to volume, Ma/Mb = 4/9 = (Aa)L/(Ab)L = Aa/Ab (Aa and Ab are the cross-sctional areass)
i.e. Aa/Ab = 4/9
Then, Ra/Rb = [pL/Aa]/[pL/Ab] = Ab/Aa = 9/4

2011-03-11 23:40:37 補充：
道理很簡單，基於長度一樣，質量與截面積成正比，但電阻值與截面積成反比。即是說，電阻值與質量成反比。於是A線與B線電阻值的比為9:4。

唔該用中文回答,唔係幾睇得明(sor開頭無講)
天同兄答案好似錯

2011-03-11 19:36:09 補充：
講真唔該用中文,睇唔明呀!!
咁次個答案正確,但係唔明囉!