Question

Apiece of wire of resistance20ω is drawn out so that its length is increased to twice its original length calculate the resistance of the wire is the new situation?

Answer 1

R=20 Ω

Let,Length of wire is L and when it drawn out, its new length is 2L

The volume of wire in both the situations remains constant.So, it the length of wire doubled, the area of cross section of wire becomes half.

So, the new resistance of wire-

Rnew = 2l/(A/2)

=4L/A

=4×Old resistance of wire (Because, R=L/A)

=4×20 = 80 Ω