A steel rod of 2 cm² area and 1 m in height is subjected to a pull of 40,000 N. If Young’s modulus is 2 × 10⁵ N/mm², the elongation of the rod in mm will be
- A10
- B100
- C1Correct
- D0.1
Solution
Concept
Calculation of axial elongation under tension
Formula
δ = (P × L) / (A × E)
Step-by-step solution
1. Given values: Load P = 40,000 N, Length L = 1 m = 1000 mm, Area A = 2 cm² = 200 mm², Young's modulus E = 2 × 10⁵ N/mm². 2. Elongation formula is ΔL = (P × L) / (A × E). 3. ΔL = (40,000 N × 1000 mm) / (200 mm² × 200,000 N/mm²) = 40,000,000 / 40,000,000 = 1 mm. Answer: C
Explanation
Elongation of a axially loaded bar is calculated using the formula where elongation equals load times length divided by the product of cross-sectional area and Young's modulus. Substituting the given values gives 40,000 N times 1,000 mm divided by 200 mm² times 200,000 N/mm². Performing the calculation yields an elongation of exactly 1 mm.
Why the other options are wrong
A) 10 mm results from an error in converting area or length units. B) 100 mm is incorrect due to a magnitude error in unit conversions. D) 0.1 mm results from incorrectly dividing the actual elongation by 10.
Shortcut
Use ΔL = PL / AE with uniform units (N, mm): ΔL = (40000 × 1000) / (200 × 200000) = 1 mm.
Common mistake
Forgetting to convert cross-sectional area from cm² to mm² or length from meters to millimeters during calculation.
Correct answer: C) 1