# A wire with a resistance R is lengthened to 1.25

discoverer02
My answer for this problem doesn't agree with the answer in the back of my textbook book:

A wire with a resistance R is lengthened to 1.25 times its original length by being pulled through a small hole. Find the resistance of the wire after it has been stretched.

R = pL/A

p = resistivity
L = length
A = area of a cross section perpendicular to the length

R1 = pL1/A1

L2 = 1.25L1 so r1/r2 should equal 1.25 right.

The decrease in radius should be proportional to the increase in length.

So r2 = r1/1.25

Therefore R2 = p(1.25L1)/[pi](r1/1.25)^2 = R1(1.25*1.56) = 1.95R1

The answer in the back of my textbook is R2 = 1.56R1

Where have I gone wrong?

Thanks

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Staff Emeritus
Gold Member
The decrease in radius should be proportional to the increase in length.

Should it? How would you go about proving this?

discoverer02
I'm assuming it's proportional. If it's not then I'm lost

Staff Emeritus
Gold Member

Originally posted by discoverer02

L2 = 1.25L1 so r1/r2 should equal 1.25 right.

Volume is constant, and volume is L*A, not L*r

discoverer02
You're right. Stupid me!

The area would have to shrink proportionally: A2 = A1/1.25 so R2 = 1.25*1.25R1 = 1.56R1.

Thank you very much for the help.