• # question_answer A wire $l=8\,m$long of uniform cross-sectional area $A=88{{m}^{2}},$ has a conductance of $G=2.45{{\Omega }^{-2}}$. The resistivity of material of the wire will be: A)  $2.1\times {{10}^{-7}}\Omega -m$ B)  $3.1\times {{10}^{-2}}\Omega -m$ C)  $4.1\times {{10}^{-7}}\Omega -m$    D)  $5.1\times {{10}^{-7}}\Omega -m$

As, the resistivity, $\rho =\frac{RA}{l}=\frac{A}{Gl}$ $=\frac{8\times {{10}^{-6}}}{2.45\times 8}=4.1\times {{10}^{-7}}\Omega m.$