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(D) The wire consists of two parallel conductors: the inner core and the outer shell.$1$. Resistance of the inner core $(R_1)$: The radius is $R$,resi

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$\frac{2}{5}\frac{\rho l}{\pi R^2}$

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(D) The wire consists of two parallel conductors: the inner core and the outer shell.$1$. Resistance of the inner core $(R_1)$: The radius is $R$,resistivity is $\rho$,and length is $l$. Thus,$R_1 = \frac{\rho l}{\pi R^2}$.$2$. Resistance of the outer shell $(R_2)$: The inner radius is $R$ and the outer radius is $2R$ (since thickness is $R$). The area of cross-section is $A_2 = \pi((2R)^2 - R^2) = 3\pi R^2$. The resistivity is $2\rho$. Thus,$R_2 = \frac{2\rho l}{3\pi R^2}$.$3$. Equivalent resistance $(R_{eq})$: Since they are in parallel,$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{\pi R^2}{\rho l} + \frac{3\pi R^2}{2\rho l} = \frac{\pi R^2}{\rho l} (1 + 1.5) = \frac{2.5 \pi R^2}{\rho l} = \frac{5 \pi R^2}{2 \rho l}$.Therefore,$R_{eq} = \frac{2 \rho l}{5 \pi R^2}$.

$A$ wire of circular cross-section has an inner portion of radius $R$ made of a material of resistivity $\rho$ and is surrounded by an outer portion of thickness $R$ made of a material of double resistivity. Find the resistance of a length $l$ of such a wire.

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