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(D) When a wire of resistance $R$ is stretched to $n$ times its original length,its volume remains constant.Since $V = A \cdot L$,if the length become

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$16 R$

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(D) When a wire of resistance $R$ is stretched to $n$ times its original length,its volume remains constant.Since $V = A \cdot L$,if the length becomes $nL$,the new area $A'$ must be $A/n$ to keep the volume constant.The resistance is given by $R = \rho \frac{L}{A}$.The new resistance $R'$ is given by $R' = \rho \frac{nL}{A/n} = n^2 \rho \frac{L}{A} = n^2 R$.Given that the length is stretched to $n = 4$ times,the new resistance is $R' = (4)^2 R = 16R$.

$A$ resistance $R$ is stretched to four times its length. Its new resistance will be

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