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(C) Let the initial length be $l$ and area be $A$. The initial resistance is $R = \rho \frac{l}{A}$.When the wire is stretched by $10\%$,the new lengt

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$1.21$ times,same

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(C) Let the initial length be $l$ and area be $A$. The initial resistance is $R = \rho \frac{l}{A}$.When the wire is stretched by $10\%$,the new length $l' = l + 0.1l = 1.1l$.Since the volume $V = Al$ remains constant,$A'l' = Al \Rightarrow A' = \frac{Al}{1.1l} = \frac{A}{1.1}$.The new resistance $R' = \rho \frac{l'}{A'} = \rho \frac{1.1l}{A/1.1} = (1.1)^2 \rho \frac{l}{A} = 1.21 R$.Specific resistance (resistivity) $\rho$ is a property of the material and depends only on the temperature,not on the dimensions of the wire.Therefore,the new resistance becomes $1.21$ times the original,and the specific resistance remains the same.

$A$ wire of a certain material is stretched slowly by $10\%$. Its new resistance and specific resistance become respectively:

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