Question

A uniform wire of resistance 100 Ω is melted and recast as a wire whose length is double that of the original. What would be the resistance of the wire?

Answer 1

Let
Resistivity of the wire = ρ
Original length of the wire = l
New length of the wire = l'
Original area of the wire = A
New area of the wire = A'
Original resistance of the wire = R = 100 Ω
New resistance of the wire = R'
Given:
l' = 2l

The volume of the wire remains constant. So,
$$\begin{gathered} Al=A\text{'}l\text{'} \\ \Rightarrow A\text{'}=\frac{A}{2} \end{gathered}$$
We know:
$$\begin{gathered} R=\frac{\rho l}{A} \\ \Rightarrow R\text{'}=\frac{\rho l\text{'}}{A\text{'}} \\ \Rightarrow \frac{R\text{'}}{R}=\frac{l\text{'}A}{lA\text{'}} \\ \because l\text{'}=2l,A\text{'}=\frac{A}{2} \\ \Rightarrow \frac{R\text{'}}{R}=\frac{\left(2l\right)A}{l\left({\displaystyle \frac{A}{2}}\right)}=4 \\ \Rightarrow R\text{'}=4R=400\mathrm{\Omega } \end{gathered}$$