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If the length of a copper plate is 10 mm and this length is increased to 15 mm after an expansion under the tensile load of 50 N. The engineering strain developed in the plate is:
1. 0.2
2. 2
3. 0.5
4. 5

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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.

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