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,

\[Al = A'l'\]

\[ \Rightarrow A' = \frac{A}{2}\]

We know:-

\[R = \frac{\rho l}{A}\]

\[ \Rightarrow R' = \frac{\rho l'}{A'}\]

\[ \Rightarrow \frac{R'}{R} = \frac{l'A}{lA'}\]

\[ \because l' = 2l,   A' = \frac{A}{2}\]

\[ \Rightarrow \frac{R'}{R} = \frac{\left( 2l \right)A}{l\left( \frac{A}{2} \right)} = 4\]

\[ \Rightarrow R' = 4R = 400  \Omega\]