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(B) The resistance of a single wire of length $l$ and diameter $d$ is given by $r = \rho \frac{l}{A} = \rho \frac{l}{\pi (d/2)^2} = \rho \frac{4l}{\pi

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$4$

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(B) The resistance of a single wire of length $l$ and diameter $d$ is given by $r = \rho \frac{l}{A} = \rho \frac{l}{\pi (d/2)^2} = \rho \frac{4l}{\pi d^2}$.When $8$ such wires are connected in parallel,the equivalent resistance $R$ is given by $R = \frac{r}{8} = \frac{1}{8} \left( \rho \frac{4l}{\pi d^2} \right) = \frac{\rho l}{2 \pi d^2}$.For a single wire of length $2l$ and diameter $d_1$ to have the same resistance $R$,we have $R = \rho \frac{2l}{\pi (d_1/2)^2} = \rho \frac{8l}{\pi d_1^2}$.Equating the two expressions for $R$: $\frac{\rho l}{2 \pi d^2} = \frac{8 \rho l}{\pi d_1^2}$.Simplifying,we get $\frac{1}{2 d^2} = \frac{8}{d_1^2}$,which implies $d_1^2 = 16 d^2$.Taking the square root,we find $d_1 = 4d$.

Eight copper wires of length $l$ and diameter $d$ are joined in parallel to form a single composite conductor of resistance $R$. If a single copper wire of length $2\,l$ has the same resistance $R$,then its diameter will be $.....d$.

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