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

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$2L$ and $d/2$

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(D) The resistance $R$ of a wire is given by the formula $R = ho \frac{L}{A}$,where $ho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.Since the volume $V$ of the copper piece remains constant,$V = A 	imes L$.Substituting $A = \frac{V}{L}$ into the resistance formula,we get $R = ho \frac{L}{V/L} = ho \frac{L^2}{V}$.Since $ho$ and $V$ are constant,$R \propto L^2$.Also,$A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.Substituting this into the resistance formula,$R = ho \frac{L}{\pi d^2 / 4} = \frac{4 ho L}{\pi d^2}$.To maximize resistance,we need to increase the length $L$ and decrease the diameter $d$.Comparing the given options,the configuration $2L$ and $d/2$ provides the maximum resistance because $R \propto \frac{L}{d^2}$.Substituting $L' = 2L$ and $d' = d/2$,we get $R' \propto \frac{2L}{(d/2)^2} = \frac{2L}{d^2/4} = 8 \frac{L}{d^2}$,which is the highest value among the choices.

$A$ piece of copper is to be shaped into a conducting wire of maximum resistance. If the initial length is $L$ and diameter is $d$,what should be the new length and diameter to achieve maximum resistance?

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