A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10^–8 Ω m. 

a) What will be the length of this wire to make its resistance 10 Ω? 

b) How much does the resistance change if the diameter is doubled?

Given a copper wire with,

Diameter, $\mathrm{d} = 0.5 \mathrm{mm} = 0.0005 \mathrm{m}$

Resistivity, $\mathrm{\rho } = 1.6 \times {10}^{–8} \Omega \mathrm{m}$

Resistance, $\mathrm{R} =10 \mathrm{\Omega }$

a) Length of the wire, $\mathrm{l} = ?$

Equation of resistance of wire,

                                                     $\mathrm{R} = \mathrm{\rho } \frac{\mathrm{l} }{\mathrm{A}}$

Area of cross-section of the wire,

                                                      $\mathrm{A}=\frac{{\mathrm{\pi d}}^2}{4}$

Hence, length of the wire,

                                                   $\mathrm{l}= \frac{\mathrm{RA}}{\mathrm{\rho }} =\frac{\mathrm{R}\times {\mathrm{\pi d}}^2}{4\mathrm{\rho }}$

Substituting the values in the above equation,

                                                $\mathrm{l}=\frac{10\times 3.14\times {\left(0.0005\right)}^2}{1.6 \times {10}^{–8}\times 4} =122.65 \mathrm{m}$

b) New resistance of the wire, ${\mathrm{R}}^{\text{'}}=?$

If the diameter of the wire is doubled, new diameter

                                          ${\mathrm{d}}^{\text{'}}=2\mathrm{d}=2\times 0.0005=1\times {10}^{-3} \mathrm{m}$

Then new area, ${\mathrm{A}}^{\text{'}}=\frac{\mathrm{\pi }{\left({\mathrm{d}}^{\text{'}}\right)}^2}{4} =\frac{4\mathrm{\pi }{\mathrm{d}}^2}{4}=4 \mathrm{A}$

Now new resistance will also be reduced 4 times.

                                                          ${\mathrm{R}}^{\text{'}}=\frac{\mathrm{R}}{4}$

                                                    $\Rightarrow {\mathrm{R}}^{\text{'}}=\frac{10}{4}=2.5 \mathrm{\Omega }$

Therefore, the length of the wire is 122.65 m and the new resistance is 2.5 Ω.