SSC JE Mechanical Engineering 2018 Set 2 Paper

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Question : 129 of 200

Marks: +1, -0

A rod of dimension 20 mm x 20 mm is carrying an axial tensile load of 10 kN. The tensile stress developed is ________.

0.025 MPa
0.25 MPa
25 MPa
250 MPa

Solution:

Given: Area

= 20 × 20 = 400 m^2

Load

= 10 KN = 10 × 10^3 N

Tensile stress =

{\Tensile \load}/{\Area}

={10×10^3}/{20×20}=25

mPa

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