Question

a given wire of resistance 1Ω is streched to double its length.What will be its new resistance?

Answer 1

R = ρL / A

R = ρL^2 / V

When wire is stretched Volume (V) remains constant.

∴ R ∝ L^2

From above relation if length (L) is doubled resistance (R) will quadruple.

Hence, new resistance will be 4Ω

Answer 2

$\tt \: \blue{Resistnace \: of \: the \: original \: wire } \\ \bf \: R = \frac{ ρL}{A}, \: \: \sf \: where \: A= \frac{V}{L} \\ \\ \bf ⟹ R = \frac{ρL²}{V} \\ \\ \cal \small { Now \: its \: length \: is \: stretched \: such \: that \: new \: length }\\ \rm L′ = 2L \\ \\ \tt \: {Thus \: resistance \: of \: the \: wire} \\ \rm \: R′ = \frac{ρ(2L)²}{V} = 4R \\ \\ \red{ \small{ \bf Hence \: \: resistance \: becomes \: 4 \: times \: the \: initial \: resistance.}}$