Question

The resistance of a wire is 'R' ohm. If it is melted and
stretched to 'n' times its original length, its new resistance
will be :- [2017]
(a)R/n
(b) n²R
(c) R/n²
(d) nR

Answer 1

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