A steel rod of 2cm diameter and 5m long is subjected to an axial pull of 3000 kg. If E = 2.1 × 106 N/mm2, the elongation of the rod will be
- A0.02275 mm
- B0.2275 mm
- C2.275 mmCorrect
- D22.275 mm
Solution
Concept
Elongation of a prismatic bar under axial load is PL/AE.
Formula
δ = P L / (A E)
Step-by-step solution
1. Take E = 2.1 × 106 kg/cm² (the unit given in the question is a typo). 2. A = π d2 / 4 = π × 22 / 4 = 3.1416 cm². 3. L = 5 m = 500 cm, P = 3000 kg. 4. δ = (3000 × 500) / (3.1416 × 2.1 × 106) = 0.2274 cm. 5. δ = 2.274 mm ≈ 2.275 mm.
Explanation
Use δ = PL/AE with consistent units in kg and cm. The elongation is about 0.2275 cm, which is 2.275 mm. In SI units (P = 29430 N, E = 2.1 × 105 N/mm²) the answer is about 2.23 mm, which is closest to option C.
Shortcut
In kg-cm units, δ = 0.2275 cm, so shift the decimal once to get mm.
Common mistake
Using E = 2.1 × 106 N/mm² literally, or mixing kg with N/mm², which gives a wrong order of magnitude.
Correct answer: C) 2.275 mm