Correct Answer - Option 2 : 24
Concept:
Strain energy stored in a bolt of length l, cross-section area A, modulus of elasticity E and is subjected to a load P is given by:
\(U = {\frac{{{P^2}l}}{{2AE}}} \)
Calculation:
Given:
EA = EB (∵ material is same), PA = PB, LA = LB, UA = 4UB, dA = 12 mm
\(\frac{{{U_A}}}{{{U_B}}} = \frac{{{{\left( {\frac{{{P^2}L}}{{2AE}}} \right)}_A}}}{{{{\left( {\frac{{{P^2}L}}{{2AE}}} \right)}_B}}} = \frac{{{A_B}}}{{{A_A}}} \)
\(4 = \frac{{d_B^2}}{{d_A^2}}\)
dB = 2 × dA = 24 mm