A wire with a resistance R is lengthened?

Question

A wire with a resistance R is lengthened to
2.91 times its original length by pulling it
through a small hole.
Find the resistance of the wire after it is
stretched. Answer in units of R.

Answer 1

If you're pulling the wire through a small hole, you can assume that the cross sectional area is changing. By conservation of volume (you're not getting rid of any of the wire) you can calculate the change in cross sectional area.

The volume of the wire is its cross sectional area times its length:

V = A * L

The lengthened wire will have the same volume, but the dimensions will change:

V = Ar * Le

where Ar is the new area and Le is the new length which is given a 2.91L. Obviously, the new area would have to be 2.91 times less than the original area to keep the volume the same:

Ar = A / 2.91

V = Ar * Le = (A / 2.91) * (2.91L) = A * L

Now we can throw in the resistance equation:

R = rho * L / A

Re = rho * Le / Ar

I have to leave you something to do, so substitute for Ar and Le in terms of A and L (and all those 2.91's), and you'll find Re in terms of R.

Answer 2

The volume of the material in the wire will not change. If length l increases by a factor of 2.91, then area A will decrease by the same factor.

R = (ρ * l) / A before
Rnew = (ρ * l*2.91) / (A/2.91) after

So the resistance increases by a factor of 2.91^2.

Answer 3

Assuming that the cross sectional area doesn't change due to the stretching of the wire, it would be 2.91R

Answer 4

if the temperature and cross section area are constant, the result is 2.91R.

if not, calculate the reductive proportion of the wire's radius from poison's ratio (not given) i.e.

σ=(Δr/r)/(ΔL/L),.

then calculate the change of area (letting x) using (Δr/r)

then the resistance'll be 2.91R/x