Question

If a wire of resistance r is melted and recast to n times its original length what is its new resistance

Answer 1

ANSWER: the resistance will Also be n times it's original resistance

because the resistance is directly proportional to length of the conductor.

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Answer 2

$\tt \: \blue{Resistnace \: of \: the \: original \: wire } \\ \bf \: R = \frac{ ρL}{A}, \: \: \sf \: where \: A= \frac{V}{L} \\ \\ \bf ⟹ R = \frac{ρL²}{V} \\ \\ \cal \small { Now \: its \: length \: is \: stretched \: such \: that \: new \: length }\\ \rm L′ = nL \\ \\ \tt \: {Thus \: resistance \: of \: the \: wire} \\ \rm \: R′ = \frac{ρ(nL)²}{V} = n²R \\ \\ \red{ \small{ \bf Hence ,\: new \: resistance \: becomes \: n² \: times \: the \: initial \: resistance.}}$