Question

The elastic limit of copper is 𝟏. 𝟓 × 𝟏𝟎𝟖 𝑵𝒎−𝟐 . If a copper wire is stretched by a load of mass 𝟖. 𝟓 𝒌𝒈 then find the minimum diameter of the wire within the elastic limit. 

Answer 1

$$p=4\frac{mg}{\pi d^2}\\1.5\cdot10^8=4\frac{8.5\cdot9.8}{\pi d^2}\\d=8.4\cdot10^{-4}\ m$$