A force of 500N1 is applied to a steel wire of cross-sectional area 0.2m2, The tensile stress is
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Correct Answer: Option D
Explanation:
\(Stress = \frac{Force}{Area}\)
\(Stress = \frac{500}{0.2} = 2500\)
= 2.5 x 103Nm-3
\(Stress = \frac{Force}{Area}\)
\(Stress = \frac{500}{0.2} = 2500\)
= 2.5 x 103Nm-3