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(B) 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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decrease $L$ and increase $D$.

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(B) 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 wire is reshaped,its volume $V = A 	imes L$ remains constant.We can express the area $A$ in terms of diameter $D$ as $A = \pi \left(\frac{D}{2}ight)^2 = \frac{\pi D^2}{4}$.Substituting $A$ into the resistance formula: $R = ho \frac{L}{\pi D^2 / 4} = \frac{4 ho L}{\pi D^2}$.Since the volume $V = A 	imes L = \frac{\pi D^2 L}{4}$ is constant,we have $L = \frac{4V}{\pi D^2}$.Substituting this into the resistance formula: $R = \frac{4 ho}{\pi D^2} 	imes \left(\frac{4V}{\pi D^2}ight) = \frac{16 ho V}{\pi^2 D^4}$.To minimize $R$,we need to maximize $D$. Since volume is constant,increasing $D$ will result in a decrease in $L$. Therefore,we should decrease $L$ and increase $D$.

$A$ copper wire of length $L$ and diameter $D$ is to be reshaped into another wire so as to have minimum resistance. For this,we should:

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