Question 16
ElasticityA force of 500N is applied to a steel wire of cross-sectional area 0.2m2. the tensile stress is
- A.1.0 × 102Nm−2
- B.1.0 × 103Nm−2
- C.2.5 × 103Nm−2
- D.2.5 × 104Nm−2
Show the answer and explanation
Answer: C (2.5 × 103Nm−2)
Explanation
Tensile stress = force ÷ area = 500 ÷ 0.2 = 2500 N m−2 = 2.5 × 103 N m−2.