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(A) Let the original length be $L_{1}$ and the original area be $A_{1}$. The original resistance is $R_{1} = \rho \frac{L_{1}}{A_{1}}$.When the length

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$9: 1$

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(A) Let the original length be $L_{1}$ and the original area be $A_{1}$. The original resistance is $R_{1} = \rho \frac{L_{1}}{A_{1}}$.When the length is increased by twice its original length,the new length $L_{2} = L_{1} + 2L_{1} = 3L_{1}$.Since the volume of the wire remains constant,$A_{1}L_{1} = A_{2}L_{2}$.Thus,$A_{2} = A_{1} \frac{L_{1}}{L_{2}} = A_{1} \frac{L_{1}}{3L_{1}} = \frac{A_{1}}{3}$.The new resistance is $R_{2} = \rho \frac{L_{2}}{A_{2}} = \rho \frac{3L_{1}}{A_{1}/3} = 9 \rho \frac{L_{1}}{A_{1}}$.Therefore,the ratio $\frac{R_{2}}{R_{1}} = \frac{9 \rho (L_{1}/A_{1})}{\rho (L_{1}/A_{1})} = 9:1$.

$A$ wire of resistance $R_{1}$ is drawn out so that its length is increased by twice of its original length. The ratio of the new resistance to the original resistance is:

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