The stress, strain and modulus of elasticity are connected by an equation. Thus, the given question has the following answers:
(i) The stress is 2 KN/[tex]mm^{2}[/tex]
(ii) The strain is 0.01
(iii) The elongation is 15 mm
From the question, given: l = 1500 mm, D = 20 mm, F = 20 000 N, λ = 2 x [tex]10^{5}[/tex] N/[tex]mm^{2}[/tex].
Thus,
(i) Stress = [tex]\frac{Force}{Area}[/tex]
 Area of the rod = Area of a circle = [tex]\pi[/tex][tex]r^{2}[/tex]
where r is the radius.
radius, r = [tex]\frac{diameter}{2}[/tex]
      = [tex]\frac{20}{2}[/tex]
r = 10 mm
so that;
Stress = [tex]\frac{20000 }{10}[/tex]
     = 2000
Stress = 2 KN/[tex]mm^{2}[/tex]
(ii) To calculate the strain;
modulus of elasticity,  λ = [tex]\frac{stress}{strain}[/tex]
2 x [tex]10^{5}[/tex] = [tex]\frac{2000}{strain}[/tex]
strain = [tex]\frac{2000}{200000}[/tex]
     = 0.01
strain = 0.01
(iii) The elongation of the rod is also its extension, e.
So that;
strain = [tex]\frac{extension}{length}[/tex]
extension = strain x length
        = 0.01 x 1500
extension  = 15 mm
Therefore: the stress is 2 KN/[tex]mm^{2}[/tex], the strain is 0.01, and the elongation is 15 mm.
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