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(C) The resistance 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-sec

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$L/2$ and $2A$

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(C) The resistance 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 mass and density of the silver are constant,the volume $V = A 	imes L$ must remain constant.We can express $A$ as $V/L$. Substituting this into the resistance formula,we get $R = ho \frac{L}{V/L} = ho \frac{L^2}{V}$.Since $ho$ and $V$ are constant,$R \propto L^2$.To minimize the resistance $R$,we must minimize the length $L$.Comparing the given options: $(a)$ $L$,$(b)$ $2L$,$(c)$ $L/2$.The smallest length is $L/2$ in option $(c)$.Therefore,the combination $L/2$ and $2A$ results in the smallest resistance.

$A$ certain piece of silver of given mass is to be made into a wire. Which of the following combinations of length $(L)$ and cross-sectional area $(A)$ will lead to the smallest resistance?

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