Question

A wire of length $L$ and radius $r$ is clamped rigidly at one end. When the other end of the wire is pulled by a force $f$, its length increases by $l$. Another wire of the same material of length $2L$ and radius $2r$, is pulled by a force $2f$. Find the increase in length of this wire.

Answer 1

Step 1: Find the change in length of the material for the first wire.

Let $Y$ be the Young's modulus of the material of the wire.

$Y=\frac{Stress}{Strain};Stress=\frac{Force}{Area}=\frac{f}{A}$;

And $Strain=\frac{Changeinlength}{Originallength}=\frac{\Delta L}{L}$

Given, For the first wire length $=L$

Applied force $=f$

Radius $=r$

Then for the first wire,

$Y=\frac{\frac{f}{\pi r^2}}{\frac{l}{L}}=\frac{fL}{\pi r^{2}l}$

$l=\frac{fL}{\pi r^{2}Y}$ ………(i)

Step 2: Find the change in length of the material for the second wire.

For the second wire length $=2L$

Applied force$=2f$

Radius $=2r$

Now the increase in length of this wire can be calculated as:

$l^{\prime }=\frac{2f\times 2L}{\pi (2r{)}^{2}Y}=\frac{fL}{\pi r^{2}Y}=l$ same as(i)

(As both the wire are made of the same material, so their Young’s modulus is same.)

Final Answer:
$l=l^{\prime }=\frac{fL}{\pi r^{2}Y}$