Correct Answer - Option 4 : 2.0 mm
Concept:
Consider prismatic bar of length L and cross-section area A is subjected to axial load P.
Then axial deformation of bar,
∆ = \(\frac{{PL}}{{AE}}\)
Calculation:
Given,
A = 100 mm2, L = 1 m = 1000 mm, Tensile force (P) = 40 KN = 40,000 N
Modulus of elasticity (E) = 200 GPa = 200 × 109 N/m2 = 200 × 109 N/mm2
So, total elongation of the rod is,
∴ ∆ = \(\frac{{PL}}{{AE}}\) = \(\frac{{40000\; \times \;1000}}{{100\; \times \;200\; \times \;{{10}^3}}}\) = 2 mm