Correct Answer - Option 4 : 0.77 mm
Concept:
Elongation due to axial load is
\(\delta l = \frac{{PL}}{{AE}}\)
And, stress \(\sigma = \frac{P}{A}\)
Calculation:
Given: L = 305 mm, σ = 276 MPa, E = 110 GPa
δl \(= \frac{{276\; \times \;305}}{{110\; \times \;{{10}^3}}}\)
∴ δl = 0.765 mm.