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(C) The resistance $R$ of a conductor is given by $R = \frac{\rho l}{A}$, where $\rho$ is the resistivity, $l$ is the length, and $A$ is the cross-sec

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$3$

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(C) The resistance $R$ of a conductor is given by $R = \frac{\rho l}{A}$, where $\rho$ is the resistivity, $l$ is the length, and $A$ is the cross-sectional area.For conductor $A$ (solid wire), the radius $r_A = 0.5 \, mm$. The cross-sectional area is $A_A = \pi r_A^2$.Thus, $R_A = \frac{\rho l}{\pi r_A^2}$.For conductor $B$ (hollow tube), the outside radius $r_{out} = 1.0 \, mm$ and the inside radius $r_{in} = 0.5 \, mm$. The cross-sectional area is $A_B = \pi (r_{out}^2 - r_{in}^2) = \pi (1.0^2 - 0.5^2) = \pi (1 - 0.25) = 0.75 \pi \, mm^2$.Comparing the areas, $A_B = \pi (1.0^2 - 0.5^2) = 3 \pi (0.5^2) = 3 A_A$.Since $R \propto \frac{1}{A}$, we have $\frac{R_A}{R_B} = \frac{A_B}{A_A} = \frac{3 A_A}{A_A} = 3$.

Two conductors are made of the same material and have the same length. Conductor $A$ is a solid wire of diameter $1.0 \, mm$. Conductor $B$ is a hollow tube of outside diameter $2.0 \, mm$ and inside diameter $1.0 \, mm$. The resistance ratio $R_A/R_B$ will be

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