What will the wire's resistance be if it is stretched to twice its original length w/o changing the volume?

Question

A wire has a resistance of 1.4×10−2 Ohm.

What will the wire's resistance be if it is stretched to twice its original length without changing the volume of the wire?

Answer 1

Underscores will indicate subscripts.

V = (pi)(r_1^2)L = (pi)(r_2^2)(2L)

(r_1)^2 = 2(r_2)^2

r_1 = 1.41 * r_2

R = Lp / A, where p is the resistivity

We've compared the 2 different L's, now let's compare the 2 different A's.

A_1 = (pi)(1.41 * r_2)^2 = (pi)(2)(r_2)^2

A_2 = (pi)(r_2)^2

A_2 = 1/2 A_1

And now we can compare the resistances:

R_1 = Lp / A

R_2 = (2L)p / (.5A) = 4Lp/A = 4 * R_1

= 4 (1.4 * 10^-2)

= 5.6 * (10^-2) Ohm, your final answer

It makes sense that the resistance should be greater, because the wire got thinner so the electrons must squeeze through a tighter area.  The wire got thinner because the length increased while the volume remained the same.  For the volume to stay the same, it had to get thinner.