[tex]\begin{gathered} R=\frac{kL}{d^2} \\ \text{For } \\ R=2.6\text{ ohm} \\ d=0.02\text{ in} \\ L=10ft=10\cdot12=120\text{ in} \\ \text{Solving K} \\ Rd^2=kL \\ k=\frac{Rd^2}{L} \\ k=\frac{(2.6ohm)(0.02in)^2}{120in} \\ k=\frac{(2.6ohm)(0.0004in^2)}{120\text{ in}} \\ k=8.67x10^{-6\text{ }}\text{ }\frac{ohm\cdot in^2}{\text{ in}} \\ \text{For } \\ d=0.01\text{ in} \\ L=\text{ 5ft=5}\cdot12=60in \\ Same\text{ wire, hence} \\ k=8.67x10^{-6\text{ }}\text{ }\frac{ohm\cdot in^2}{\text{ in}} \\ R=\frac{(8.67x10^{-6\text{ }}\text{ }\frac{ohm\cdot in^2}{\text{ in}})(60\text{ in)}}{(0.01\text{ in})^2} \\ R=5.202\text{ ohm} \\ The\text{ new resistance is }5.202\text{ ohm} \end{gathered}[/tex]
