SSC JE Mechanical Engineering 2018 Set 5 Paper

Question : 119 of 200

Marks: +1, -0

A steel rod of original length 200mm and final length of 200.2 mm after application of an axial tensile load of 20 kN, what will be the strain developed in the rod?

0.01
0.1
0.001
0.0001

Validate

Solution:

Strain

={\change \in \length}/{\original \length}={200.2- 200}/200

Strain = 0.001

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