Answer
(a) $1.8\times 10^{-3}$
(b) $0.60 cm$
Work Step by Step
(a) We know that
$F=Y(\frac{\Delta L}{L_{\circ}})A$
This can be rearranged as:
$\frac{\Delta L}{L_{\circ}}=\frac{F}{YA}$
We plug in the known values to obtain:
$\frac{\Delta L}{L_{\circ}}=\frac{2.548\times 10^{-3}N}{(4.7\times 10^9N/m^2)(3.016\times 10^{-10}m^2)}$
$\frac{\Delta L}{L_{\circ}}=1.8\times 10^{-3}$
(b) We can find the required radius as follows:
$r=\sqrt{\frac{F}{\pi Y(\frac{\Delta L}{L_{\circ}})}}$
We plug in the known values to obtain:
$r=\sqrt{\frac{744.8N}{\pi(0.37\times 10^{10})(1.8\times 10^{-3})}}$
$r=5.967\times 10^{-3}m=0.60cm$
