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(B) Given: Length $L = 100\,cm = 1\,m$,Diameter $d = 2.0\,mm = 2.0 	imes 10^{-3}\,m$,Resistance $R = 0.7\,\Omega$.Radius $r = \frac{d}{2} = 1.0 	ime

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$2.2$

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(B) Given: Length $L = 100\,cm = 1\,m$,Diameter $d = 2.0\,mm = 2.0 	imes 10^{-3}\,m$,Resistance $R = 0.7\,\Omega$.Radius $r = \frac{d}{2} = 1.0 	imes 10^{-3}\,m$.The cross-sectional area $A = \pi r^2 = \frac{22}{7} 	imes (1.0 	imes 10^{-3})^2 = \frac{22}{7} 	imes 10^{-6}\,m^2$.Using the formula $R = \frac{ho L}{A}$,we get $ho = \frac{R \cdot A}{L}$.Substituting the values: $ho = \frac{0.7 	imes (\frac{22}{7} 	imes 10^{-6})}{1}$.$ho = 0.1 	imes 22 	imes 10^{-6} = 2.2 	imes 10^{-6}\,\Omega \cdot m$.Thus,the resistivity is $2.2 	imes 10^{-6}\,\Omega \cdot m$.

$A$ wire $100\,cm$ long and $2.0\,mm$ diameter has a resistance of $0.7\,\Omega$. The electrical resistivity of the material is ...........$ \times 10^{-6}\,\Omega \cdot m$.

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