Question

A uniform wire of resistance 100 $\Omega$ is melted and recast in a wire of length double that of the original. What would be the resistance of the wire ?

Answer 1

$l=2l\text{Volume of wire remains constant}$

$Ai=A^{\prime }{i}^{\prime }$

$\Rightarrow Ai=A^{\prime }\times 2i$

$\Rightarrow A^{\prime }=\frac{A}{2}$

$p\Rightarrow \text{Specific resistance}$

$R=\frac{pl}{A^{\prime }}$

$R_1=\frac{pl}{A}=100\Omega$

$=\frac{p2l}{\frac{A}{2}}=\frac{4pl}{A}=4R$

$=4\times 100\Omega =400\Omega$