Question

Find the extension produced in a copper wire of length 2 m and diameter 3 mm when a load of 30 N is applied. (Young’s modulus for copper = 11 x 10^11 Nm-2)

Answer 1

Mechanical stress:

$\sigma=\frac{F}{S}=\epsilon E$

where $F$ - is a force of the load.

$S=\frac{\pi d^2}{4}$

$\epsilon=\frac{\Delta l}{L_0}$

$E$ - Young's module.

We can rewrite this formula as

$\frac{4F}{\pi d^2}=\frac{\Delta l}{L_0}E$

hence

$\Delta l=\frac{4FE}{\pi d^2}l_0$