Question

The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be ______%.

Answer 1

Volume is constant so on length doubled

Area is halfed so

$R=p\frac{l}{A}$

and

$R'=p\frac{2l}{\frac{A}{2}}=4p\frac{l}{A}=4R$

So percentage increase will be

R%$=\frac{4R-R}{r}×100=300$