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(B) Given: Current $I = 2\,A$,Potential difference $V = 3.4\,V$,Mass $m = 8.92 	imes 10^{-3}\,kg$,Density $d = 8.92 	imes 10^3\,kg/m^3$,Resistivity

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$10$

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(B) Given: Current $I = 2\,A$,Potential difference $V = 3.4\,V$,Mass $m = 8.92 	imes 10^{-3}\,kg$,Density $d = 8.92 	imes 10^3\,kg/m^3$,Resistivity $ho = 1.7 	imes 10^{-8}\,\Omega\cdot m$.Using Ohm's Law,the resistance $R$ is given by $R = \frac{V}{I} = \frac{3.4}{2} = 1.7\,\Omega$.We know that $R = ho \frac{l}{A}$,where $l$ is the length and $A$ is the cross-sectional area.Also,the volume $V_{vol} = A \cdot l = \frac{m}{d} = \frac{8.92 	imes 10^{-3}}{8.92 	imes 10^3} = 10^{-6}\,m^3$.Thus,$A = \frac{10^{-6}}{l}$.Substituting $A$ into the resistance formula: $1.7 = ho \frac{l}{(10^{-6}/l)} = ho \frac{l^2}{10^{-6}}$.$l^2 = \frac{1.7 	imes 10^{-6}}{ho} = \frac{1.7 	imes 10^{-6}}{1.7 	imes 10^{-8}} = 10^2$.Therefore,$l = \sqrt{100} = 10\,m$.

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