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

Answer

`80 Omega`

Explanation

If the original length and the cross-sectional area of the wire are `l and A`, respectively, then
`R = (rho l)/(A)` or `20 Omega = (rho l)/(A)`.
When the length becomes `2 l` and cross-sectional area is `A//2`.
`R_("new") = (rho(2 l))/(A//2) = 4(rho l//A) = 4(20 Omega) = 80 Omega`.