Answer
$r=3.5\times10^{-4}m$
Work Step by Step
$A=\frac{Fl_o}{E\Delta l}$
The upward force of tension of the two strings balance the downward force of the weight of the mass.
$2F_T\sin(12^o)=245N$
$F_T=589N$
$x$ is the length of half of the string before mass is attached.
$\frac{x}{\cos(12)}$ is the length of half of the string after mass is attached.
$\Delta l=\frac{x}{\cos(12)}-x=x(\frac{1}{\cos(12)}-1)$
$\frac{l_o}{\Delta l}=\frac{x}{x(\frac{1}{\cos(12)}-1)}=\frac{\cos(12)}{1-\cos(12)}=44.8$
$E=70\times10^9N/m^2$
Plugging them all into the equation,
$A=\frac{(589N)(44.8)}{70\times10^9N/m^2}=3.766\times10^{-7}m^2$
$r=3.5\times10^{-4}m$
