Change the chapter
Suppose the resistance of a wire is 2 Ω. If the wire is stretched to three times its length, what will be its resistance? Assume that the volume does not change.
Question by OpenStax is licensed under CC BY 4.0.
Final Answer

$18 \Omega$

Solution Video

OpenStax College Physics for AP® Courses Solution, Chapter 20, Problem 13 (Test Prep for AP® Courses) (2:55)

Sign up to view this solution video!

View sample solution
Video Transcript

This is College Physics Answers with Shaun Dychko. Resistance in the first case will be the resistivity multiplied by the first length divided by the first cross-sectional area. We can replace <i>A one</i> with pi times radius one squared. Then we consider the second case where the wire is stretched and it's going to be resistivity which does not need a subscript because it's the same material in both cases, times the new length <i>l two</i> divided by pi times the new radius <i>r two</i> squared. There's a new radius because as the wire is stretched out it's also going to get more thin because the volume is not changing. So it's the same amount of material in both cases but in case two, it's just longer and so it has to have a compensating reduction in its radius in order to extend its length. Now we know that <i>l two</i> is three times <i>l one</i> we're told, it's stretched to three times its length, but we don't know what <i>r two</i> is yet. So let's consider volumes in order to make a relation between <i>rone</i>squared and <i>r two</i> squared because at this point we can't figure out what <i>r two</i> is in terms of <i>r one</i>. So we know the volume in the first case is pi times <i>r one</i> squared times <i>l one</i>, this is the volume of a cylinder, the cross-sectional area multiplied by its length. In the second case it's pi times radius two squared times <i>l two</i>. Replacing <i>l two</i> with three times <i>l one</i>, we get three pi <i>r two</i> squared times <i>l one</i> is the volume in case two. We're told that the volume doesn't change, <i>v one</i> equals <i>v two</i>. So I didn't even need a subscript on the v's there. So now we can say that this which is here equals volume two, three pi <i>r two</i> squared times <i>l one</i>. Then we can divide both sides by three pi <i>l one</i> and we end up with <i>r two</i>squared equals <i>r one</i> squared over three. I won't bother taking the square root of both sides because it's <i>r two</i> squared that we care about anyway because that's what's showing up in our formula. So we're going to substitute now in the <i>r two</i> formula with radius two squared. We're going to multiply by its reciprocal because dividing by this thing is the same as multiplying by its reciprocal. So we're going to multiply by three over <i>r one</i> squared, and we end up with three times three making nine times <i>rho</i> times <i>l one</i> over pi <i>r one</i> squared. This is resistance one as we showed up here. So resistance two is going to be nine times resistance one. We're told resistance one is two ohms and so resistance two is 18 ohms.