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

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One-fourth

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(D) The resistance $R$ of a wire is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is resistivity,$L$ is length,and $A$ is the cross-sectional area.Since the wire is cylindrical,the cross-sectional area $A = \pi r^2$,where $r$ is the radius.Since the diameter $D = 2r$,we can write $A = \pi (D/2)^2 = \frac{\pi D^2}{4}$.Substituting this into the resistance formula: $R = \rho \frac{L}{\pi D^2 / 4} = \frac{4 \rho L}{\pi D^2}$.This shows that $R \propto \frac{1}{D^2}$.If the diameter $D$ is doubled $(D' = 2D)$,the new resistance $R'$ becomes $R' = \frac{4 \rho L}{\pi (2D)^2} = \frac{4 \rho L}{4 \pi D^2} = \frac{1}{4} R$.Therefore,the resistance becomes one-fourth of its original value.

If the diameter of a conducting wire is doubled,how many times does its resistance become?

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