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

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$\frac{R}{n^4}$

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(A) The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $\rho$ is resistivity,$L$ is length,and $A$ is the cross-sectional area. Since $A = \pi r^2$,we have $R = \rho \frac{L}{\pi r^2}$.Since the volume $V = A \cdot L$ remains constant during compression,$L_1 A_1 = L_2 A_2$.Given $A_2 = \pi (nr)^2 = n^2 A_1$,we find $L_2 = L_1 \frac{A_1}{A_2} = \frac{L_1}{n^2}$.The new resistance $R_2$ is $R_2 = \rho \frac{L_2}{A_2} = \rho \frac{L_1 / n^2}{n^2 A_1} = \frac{1}{n^4} \left( \rho \frac{L_1}{A_1} \right) = \frac{R}{n^4}$.

$A$ uniform wire of resistance $R$ is uniformly compressed along its length,until its radius becomes $n$ times the original radius. Now,the resistance of the wire becomes:

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