Stretching a Thin Wire

The wire's volume is equal to $\pi A l$ , where $A$ is the cross-sectional area of the wire, and $l$ is its length. When the wire is stretched, we are told that the volume remains constant. We know that $l$ is now $1.4l$ . Let us call the new cross-sectional area of the wire $A_n$ . Then $\pi A l = \pi A_n \times 1.4l$ $A = 1.4 A_n$ $A_n = \frac{A}{1.4}$ Resistance is directly proportional to length, so the increased length increases $R$ by a factor of $1.4$ . Resistance is inversely proportional to cross-sectional area, so the decreased area increases $R$ by a factor of $1/(1/1.4) = 1.4$ .

Therefore $R$ has been increased by a factor of $1.4 \times 1.4 = 1.96$ and so the answer is $96$ %.