\(r = {d \over 2} = 2,{5.10^{ - 2}}(m);E = {7.10^{10}}(Pa);F = 3450N\). Tính \({{\Delta l} \over {{l_0}}}\)
\(F = {{ES} \over {{l_0}}}.\Delta l\)
\(\Rightarrow \varepsilon = {{\Delta l} \over {{l_0}}} = {F \over {ES}} = {{3450} \over {{{7.10}^{10}}.3,14.{{(2,{{5.10}^{ - 2}})}^2}}} = 0,0025(\% )\)