Chapter 17 - Electric Current and Resistance - Learning Path Questions and Exercises - Exercises - Page 620: 22

a). $2.5\times10^{-5}ohm-m$ b). Semiconductor.

a). Area = $a^{2}=(0.5\times10^{-2})^{2}m^{2}=2.5\times10^{-5}m^{2}$ Now, $R=V/I=100V/5A=20ohms$ Since, $R=p\frac{L}{A}$ $resistivity = p=\frac{RA}{L}=\frac{20\times2.5\times 10^{-5}}{20}=2.5\times10^{-5}ohm-m$ b). If resistivity is in the order of $10^{-5}$, then it is a semiconductor. So the given material is a semiconductor.