A cylindrical copper rod has length L and resistance R. If it is melted and formed into another rod of length 2L, the resistance will be

${\text{L}}_{\text{1}}=\text{L},{\text{R}}_{\text{1}}=\text{R}$ 

${\text{L}}_{\text{2}}=2\text{L},{\text{R}}_{\text{2}}=\text{?}$

          $l_1=L$

${\text{A}}_{\text{1}}=\text{A}$

$\text{V}={\text{A}}_{\text{1}}{\text{L}}_{\text{1}}$

$l_2=\text{2L}$

${\text{A}}_{\text{2}}=\text{A/2}$

$\text{V}={\text{A}}_{\text{2}}{\text{L}}_{\text{2}}$

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

${\text{R}}_1=\frac{\rho l}{\text{A}}$

${\text{R}}_{\text{2}}=\frac{\rho \left(2l\right)}{\text{A/2}}=4\text{R}$

Volume should remain the same

$\frac{{\text{R}}_{\text{1}}}{{\text{R}}_{\text{2}}}=\frac{l_1{ }^2}{l_2{ }^2}=\frac{{\text{A}}_{\text{1}}{ }^{\text{2}}}{{\text{A}}_{\text{2}}{ }^{\text{2}}}$