For a circular cross-section:
\( A = \pi r^{2} = \pi (0.015)^{2} = \pi \times 2.25 \times 10^{-4} \,\text{m}^{2} \approx 7.07 \times 10^{-4} \,\text{m}^{2} \)
Stress is force per unit area:
\( \sigma_{\text{max}} = \dfrac{F_{\text{max}}}{A} \Rightarrow F_{\text{max}} = \sigma_{\text{max}} \, A \)
\( F_{\text{max}} = 10^{8} \times 7.07 \times 10^{-4} \,\text{N} = 7.07 \times 10^{4} \,\text{N} \)
Maximum load the cable can support: \( F_{\text{max}} \approx 7.1 \times 10^{4} \,\text{N} \) (about \( 71 \,\text{kN} \)).


The Young's modulus of a steel wire of length \(6 m\) and cross-sectional area \(3 \,mm ^2\), is \(2 \times 10^{11}\) \(N / m ^2\). The wire is suspended from its support on a given planet A block of mass \(4 kg\) is attached to the free end of the wire. The acceleration due to gravity on the planet is \(\frac{1}{4}\) of its value on the earth The elongation of wire is (Take \(g\) on the earth \(=10\, m / s ^2\)) :