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A steel of cross-section area 100 mm2 and 1 m long is subjected to a tensile force of 40 kN. What is the total elongation of the rod? If modulus of elasticity of steel is 200 GPa.
1. 0.5 mm
2. 0.7 mm
3. 1.2 mm
4. 2.0 mm

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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

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