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

Explanation

Suppose the length of `R_(1)= 20Omega` resistance wire is l, its area of cross-section is A and its resistivity is `rho`. Then
`R_(1)= (rhol)/(A)= 20Omega`
When length becomes twice (2l), then its area of cross-section will become half `((A)/(2))` The new resistance
`R_(2)= (rho.2l)/(A//2)= R_(2)= 4(rhol)/(A)= 4 xx 20= 80Omega`