Question

If the elastic limit of copper is $1.5\times {10}^{8}N/m^8$, the minimum diameter a copper wire can have under a load of $10.0kg$, if its elastic limit is not to be exceeded, is $\frac{x}{10}mm$. Find x.

Answer 1

Let $r$ be the radius of the copper wire.
Elastic limit is the breaking stress $\sigma =\frac{F}{A}=\frac{mg}{\pi r^2}$
$\therefore r=\sqrt{\frac{mg}{\pi \sigma }}=\sqrt{\frac{10\times 9.8}{3.14\times 1.5\times {10}^8}}=4.56\times {10}^{-4}m$
Diameter of the wire is $d=2r=0.91\times {10}^{-3}m=0.9mm=\frac{9}{10}mm$