A uniform wire having a resistance of 100 is melted and recast into a new wire whose length is double that of the original. What will be the resistance of the new wire?

Step-by-step solution

Let the original resistance be R₁ = 100 , original length be L₁ , and original area of cross-section be A₁ . The resistance of the original wire is given by R₁ = L₁ A₁ . When the wire is melted and recast, its volume remains constant. Thus, V = L₁ A₁ = L₂ A₂ . Given that the new length is L₂ = 2 L₁ . The new area of cross-section is A₂ = V L₂ = L₁ A₁ 2 L₁ = A₁ 2 . The resistance of the new wire is R₂ = L₂ A₂ = 2 L₁ A₁ 2 = 4 L₁ A₁ . Therefore, R₂ = 4 R₁ = 4 100 = 400 .