A piece of wire of resistance 20Ω is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new situation.

We know that, R = ρl/A Given that,

l’ = 2l A’ = A/2

And ρ’ = ρ (as the material of the wire is the same) Therefore,

R’ = ρ’l’/A’

= ρ2l/(A/2)

= 4ρl/A

= 4R

Therefore, R’ = 4×20 = 80Ω