Correct Answer - Option 3 : 0.5
Concept:
Engineering strain (ϵ),
\(ϵ=\frac{\delta L}{L_i}= \frac{(L_f-L_i)}{L_i} \)
Calculation:
Given:
Li = 10 mm, Lf = 15 mm.
\(ϵ=\frac{\delta L}{L_i}= \frac{(L_f-L_i)}{L_i} \)
\(ϵ= \frac{(15\;-\;10)}{10} = 0.5 \)
∴ The engineering strain developed in the plate is 0.5.