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A copper piece originally 305 mm long is pulled in tension with a stress of 276 MPa. If the deformation is entirely elastic and the modulus of elasticity is 110 GPa, the resultant elongation will be nearly
1. 0.43 mm
2. 0.54 mm
3. 0.65 mm
4. 0.77 mm

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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.

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