Answer
(a) $0.11\frac{\Omega}{m}$
(b) decrease
(c) $0.074\frac{\Omega}{m}$
Work Step by Step
(a) We know that
$R=\frac{\rho L}{A}$
This can be rearranged as:
$\frac{R}{L}=\frac{\rho}{A}$
We plug in the known values to obtain:
$\frac{R}{L}=\frac{2.8\times 10^{-8}}{2.4\times 10^{-7}}=0.11\frac{\Omega}{m}$
(b) As We know that $\frac{R}{L}=\frac{\rho}{A}$
The above equation shows that resistance per meter is inversely proportional to the cross sectional area; so by increasing the diameter, the cross sectional area increases and hence resistance per meter decreases. Thus our answer to part(a) decreases.
(c) We know that
$R=\frac{\rho L}{A}$
This can be rearranged as:
$\frac{R}{L}=\frac{\rho}{A}$
We plug in the known values to obtain:
$\frac{R}{L}=\frac{2.8\times 10^{-8}}{3.6\times 10^{-7}}=0.074\frac{\Omega}{m}$
