Answer:
105.70 mm
Explanation:
Poisson’s ratio, v is the ratio of lateral strain to axial strain.
E=2G(1+v) where E is Young’s modulus, v is poisson’s ratio and G is shear modulus
Since G is given as 25.4GPa, E is 65.5GPa, we substitute into our equation to obtain poisson’s ratio
[tex]\begin{array}{l}\\65.5{\rm{ GPa}} = 2\left( {25.4{\rm{ GPa}}} \right)(1 + \upsilon )\\\\\upsilon = 0.2893\\\end{array}[/tex]
Original length [tex]L_(i}[/tex]
[tex]\upsilon = - \left( {\frac{{\left( {\frac{{{d_f} - {d_i}}}{{{d_i}}}} \right)}}{{\left( {\frac{{{L_f} - {L_i}}}{{{L_i}}}} \right)}}} \right)[/tex]
Where [tex]d_{f}[/tex] is final diameter, [tex]d_{i}[/tex] is original diameter, [tex]L_{f}[/tex] is final length and [tex]L_{i}[/tex] is original length.
[tex]\begin{array}{l}\\0.2893 = - \left( {\frac{{\left( {\frac{{30.04{\rm{ mm}} - {\rm{30 mm}}}}{{{\rm{30 mm}}}}} \right)}}{{\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right)}}} \right)\\\\\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right) = - 4.6088 \times {10^{ - 3}}\\\end{array}[/tex]
[tex]\begin{array}{l}\\105.2 - {L_i} = - \left( {4.6088 \times {{10}^{ - 3}}} \right){L_i}\\\\105.2 = 0.9953{L_i}\\\\{L_i} = 105.70{\rm{ mm}}\\\end{array}[/tex]
Therefore, the original length is 105.70 mm
