A uniform heavy rod of mass 20 kg, cross-sectional area 0.4 m² and length 20 m is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is x×10⁻⁹ m. The value of x is . (Given: Young's modulus Y=2×10¹¹ N m⁻² and g=10 m s⁻²)

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25. Idea: the tension falls linearly from mg at the support to zero at the free end, so the effective load is the average, (mg)/2. Δ l=(mgL)/(2AY), the self-weight result — half of what the same weight hung at the end would produce. Given: mg=20×10=200 N, L=20 m, A=0.4 m², Y=2×10¹¹ N m⁻². Δ l=(200×20)/(2×0.4×2×10¹¹)=(4000)/(1.6×10¹¹)=2.5×10⁻⁸ m. 2.5×10⁻⁸=25×10⁻⁹, so x=25.

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