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(C) Resistance of a conductor is given by $R = \frac{\rho l}{A}$,where $\rho$ is resistivity,$l$ is length,and $A$ is the cross-sectional area.For con

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$3:1$

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(C) Resistance of a conductor is given by $R = \frac{ho l}{A}$,where $ho$ is resistivity,$l$ is length,and $A$ is the cross-sectional area.For conductor $A$ (solid wire of diameter $d_A = 1 \ mm$):Radius $r_A = 0.5 \ mm = 0.5 	imes 10^{-3} \ m$.Area $A_A = \pi r_A^2 = \pi (0.5 	imes 10^{-3})^2 = 0.25 \pi 	imes 10^{-6} \ m^2$.$R_A = \frac{ho l}{0.25 \pi 	imes 10^{-6}} = \frac{4 ho l}{\pi 	imes 10^{-6}}$....$(1)$For conductor $B$ (hollow tube of outer diameter $D_B = 2 \ mm$ and inner diameter $d_B = 1 \ mm$):Outer radius $R_{out} = 1 \ mm = 10^{-3} \ m$,inner radius $r_{in} = 0.5 \ mm = 0.5 	imes 10^{-3} \ m$.Area $A_B = \pi (R_{out}^2 - r_{in}^2) = \pi ((10^{-3})^2 - (0.5 	imes 10^{-3})^2) = \pi (1 - 0.25) 	imes 10^{-6} = 0.75 \pi 	imes 10^{-6} \ m^2$.$R_B = \frac{ho l}{0.75 \pi 	imes 10^{-6}} = \frac{4 ho l}{3 \pi 	imes 10^{-6}}$....$(2)$Taking the ratio $\frac{R_A}{R_B}$:$\frac{R_A}{R_B} = \frac{4 ho l}{\pi 	imes 10^{-6}} 	imes \frac{3 \pi 	imes 10^{-6}}{4 ho l} = \frac{3}{1}$.Thus,$R_A : R_B = 3 : 1$.

Two conductors are made of the same material and have the same length. Conductor $A$ is a solid wire of diameter $1 \ mm$. Conductor $B$ is a hollow tube of outer diameter $2 \ mm$ and inner diameter $1 \ mm$. Find the ratio of resistance $R_A$ to $R_B$.

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