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(B) The resistance 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 a

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$0.25$

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(B) The resistance 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 area $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$,we have $R \propto \frac{1}{d^2}$.Let the original resistance be $R_1 = R$ with diameter $d_1 = d$.For the second wire,the diameter is $d_2 = 2d$.Therefore,the new resistance $R_2$ is given by $\frac{R_2}{R_1} = \frac{d_1^2}{d_2^2} = \frac{d^2}{(2d)^2} = \frac{d^2}{4d^2} = \frac{1}{4}$.Thus,$R_2 = \frac{R}{4} = 0.25\,R$.

$A$ certain wire has a resistance $R$. The resistance of another wire identical with the first except having twice its diameter is (in $,R$)

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