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(B) The resistance of a wire is given by $R = \rho \frac{L}{A}$, where $\rho$ is the resistivity, $L$ is the length, and $A$ is the cross-sectional ar

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$n^2 R$

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(B) The resistance of a wire is given by $R = ho \frac{L}{A}$, where $ho$ is the resistivity, $L$ is the length, and $A$ is the cross-sectional area.When the wire is elongated $n$-fold, the new length $L' = nL$.Since the volume of the wire remains constant, $V = A 	imes L = A' 	imes L'$.Therefore, $A' = \frac{A 	imes L}{L'} = \frac{A 	imes L}{nL} = \frac{A}{n}$.The new resistance $R'$ is given by $R' = ho \frac{L'}{A'} = ho \frac{nL}{A/n} = n^2 \left( ho \frac{L}{A} ight) = n^2 R$.

$A$ wire of resistance $R$ is elongated $n$-fold to make a new uniform wire. What is the resistance of the new wire?

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