Question

A wire of resistance r is stretched till its length becomes n times..What is its new resistance

Answer 1

The resistance of a uniform wire is directly proportional to its length and inversely proportional to its cross-sectional area.

Here, R is the resistance of the wire in ohms, l is the length of the wire in meters, and A is the cross-sectional area of the wire in.

If the length of the wire is stretched uniformly to 2 times its original length, then the cross-sectional area is reduced to times because the volume remains the same.

Since the resistivity of the material constant, thus calculate the resistance of the wire.

The resistance of the wire is 4 times higher to its original resistance.

Answer 2

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