
$\mathrm{~A}=\pi \mathrm{r}^2=16 \pi \times 10^{-10} \mathrm{~m}^2$
$\text { Strain }=\frac{\Delta \ell}{\ell}=\frac{\mathrm{F}}{\mathrm{AY}}=\frac{\mathrm{T}}{\mathrm{AY}}$
$=\frac{80 / 3}{16 \pi \times 10^{-10} \times 2 \times 10^{11}}=\frac{1}{12 \pi}$
$\alpha=12$
