Question

A wire of resistance r is stretched to double its length its new resistance is

Answer 1

Answer:-

New resistance = 8 × original resistance.

Given:-

length of wire is doubled.

To find :-

The new resistance.

Solution :-

Let the length be l.

Let the length be l.Area be A

Resistance be R.

When it is not stretched :-

The resistance is given by :-

$R = \rho \dfrac{l}{A}$

After stretching the wire :-

After stretching the wire :- L" = 2l

After stretching the wire :- L" = 2l

Area" = 1/4 A

Now,resistance is :-

$R" = \rho \dfrac {l"}{A"}$

put the value

$R" = \rho \dfrac {2l}{\dfrac{A}{4}}$

$R" = \rho \dfrac{8l}{A}$

$R" = 8R$

hence,the resistance of wire increases 8 times than original value.

Answer 2

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