Answer
$5.7\times10^{2}\,N$
Work Step by Step
Given that stress is the same for both the cables, this implies that
$\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}$
$F_{1}=270\,N$
$A_{1}=\pi r_{1}^{2}=\pi(3.5\times10^{-3}\,m)^{2}$
$A_{2}=\pi r_{2}^{2}=\pi (5.1\times10^{-3}\,m)^{2}$
$F_{2}=\frac{F_{1}}{A_{1}}\times A_{2}=\frac{270\,N}{\pi(3.5\times10^{-3}\,m)^{2}}\times\pi (5.1\times10^{-3}\,m)^{2}=5.7\times10^{2}\,N$
