Question:
An iron rod of length 2 m and cross-sectional area 50 mm² is stretched by 0.5 mm by hanging a mass of 250 kg. The Young’s modulus of iron is:
An iron rod of length 2 m and cross-sectional area 50 mm² is stretched by 0.5 mm by hanging a mass of 250 kg. The Young’s modulus of iron is:
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Convert mm² to m² carefully.
Updated On: Mar 18, 2026
- 19.6×10¹⁰ N/m²
- 19.6×10¹⁸ N/m²
- 19.6×10¹¹ N/m²
- 19.6×10¹⁵ N/m²
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Formula.
Y = (FL)/(AΔ L)
Step 2: Substitution.
Gives 19.6×10¹¹ N/m².
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