Question

A copper wire of length l and radius r has a resistance r.This wire is stretched to make its radius half.Find its new resistance

Answer 1

given length l
radius r
Resistance R
Resistance Becomes
$r = \gamma( l \div \pi {r}^{2} )$
radius becomes half then Resistance
R(new) = 4 R

Answer 2

Answer:

The new resistance will be 16 times of the old resistance.

Explanation:

Given that,

Length = l

Radius = r

Resistance = R

This wire is stretched to make its radius half.

The volume will be same

$V_{1}=V_{2}$

$\pi r_{1}^2 l_{1}=\pi r_{2}^{2} l_{2}$

$\pi r^2 l_{1}=\pi (\dfrac{r}{2})^2 l_{2}$

$l_{1}=\dfrac{l_{2}}{4}$

Therefore,

$l_{1}=l$

$l_{2}=4 l$

$r_{1}=r$

$r_{2}=\dfrac{r}{2}$

The resistance is defined as:

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

$R= \dfrac{\rho l}{\pi r^2}$

Here, $\rho$= resistivity of the wire

l= length of the wire

A= Area of cross section

The new resistance will be

$R'=\dfrac{\rho l}{\pi r_{2}^2}$

The ratio of the old resistance and new resistance

$\dfrac{R}{R'}=\dfrac{\dfrac{\rho l_{1}}{\pi r_{1}^2}}{\dfrac{\rho l_{2}}{\pi r_{2}^2}}$

$\dfrac{R}{R'}=\dfrac{l_{1}\times r_{2}^2}{l_{2}\times r_{1}^2}$

$\dfrac{R}{R'}=\dfrac{l\times \dfrac{r^2}{4}}{4l\times r^2}$

$\dfrac{R}{R'}=\dfrac{1}{16}$

$R'=16 R$

Hence, The new resistance will be 16 times of the old resistance.