A wire of resistance `5Omega` is drawn out so that its length is increased by twice its original length. Calculate its new resistance.

Explanation

The resistance of a wire of length l and radius r is given by `R_(1) = rho (l)/(pi r^(2))` [ `R_(1) = 5Omega` (Given)]
When its length is increased to twice its original length, the radius of wire becomes `(r )/(2)`
`:. R = rho (2l)/((r )/(2)) =4 (rho (1)/(r ))^(2) = 4R_(1)`
or `R= 4 xx 5 = 20 Omega`