Chapter 17 - Electric Current and Resistance - Learning Path Questions and Exercises - Exercises - Page 620: 21

$1.3\times10^{-2} Ω$

Length of the wire l= 0.60 m Diameter d= 0.10 cm Radius r= d/2 = 0.05 cm= $0.05\times10^{-2}m$ Area of cross section A=$\pi r^{2}=\pi(0.05\times10^{-2}m)^{2}$ Resistivity of copper $\rho= 1.70\times10^{-8}Ωm$ We know that resistance $R= \frac{\rho l}{A}$. Substituting the values in this equation, we get $R= \frac{1.70\times10^{-8}Ωm\times0.60m}{\pi(0.05\times10^{-2}m)^{2}}=1.3\times10^{-2} Ω$