Elongation due to self weight,
2AYWl=2×2×1011×100×10−410×0.2=5×10−10m
A uniform heavy rod of weight 10kgms−2, cross-sectional area 100cm2 and length 20cm is hanging from a fixed support. Young modulus of the material of the rod is 2×1011Nm−2. Neglecting the lateral contraction, find the elongation of rod due to its own weight:
Held on 31 Aug 2021 · Verified 6 Jul 2026.
5×10−10m
4×10−8m
5×10−8m
2×10−9m
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