A force of 500 N is applied to one end of a cylindrical steel rod of diameter 50cm, the tensile stress is;
Correct answer: C. 2.5 x 10^3Nm- 2
- A. 1.5 x 10^5Nm- 2
- B. 2.5 x 10^5Nm- 2
- C. 2.5 x 10^3Nm- 2
- D. 1 x 10^5Nm- 2
Explanation
We use the formula: σ= F/A Given:Force (F) = 500 NDiameter of the rod = 50 cm = 0.5 mThe cross-sectional area of a cylinder is calculated using the formula for the area of a circle ( A = πr2 ), where ( r ) is the radius of the circle.The radius is half of the diameter, so:r =0.5m/2 =0.25mNow, we can calculate the area:A = π(0.25)2 = π(0.0625 = 0.19635m² (rounded to 5 decimal places)Finally, we can calculate the tensile stress: σ = 500/0.19635 = 2546.48 Nm-2So, the tensile stress is approximately 2546.48 or 2.5×103Nm- 2.
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