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Elongation of a uniform rod hanging under its own weight

This type of question, asked 2 times: 2021 Paper 22, 26 Jul 2022. All 2 questions below, oldest first.

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  1. Question 1

    2021 Paper 22

    12
    A uniform heavy rod of weight 10 kg ms−210\text{ }kg{\text{ }ms}^{- 2}, cross-sectional area 100 cm2100{\text{ }cm}^{2} and length 20 cm20\text{ }cm is hanging from a fixed support. Young modulus of the material of the rod is 2×1011Nm−22 \times10^{11}Nm^{- 2}. Neglecting the lateral contraction, find the elongation of rod due to its own weight.

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  2. Question 2

    26 Jul 2022

    115
    A uniform heavy rod of mass 20 kg20\text{ }kg, cross-sectional area 0.4 m20.4{\text{ }m}^{2} and length 20 m20\text{ }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 mx \times10^{- 9}\text{ }m. The value of xx is :(Given. Young's modulus Y=2×1011Nm−2Y = 2 \times10^{11}Nm^{- 2} and  g=10 ms−2)\left. \ g = 10{\text{ }ms}^{- 2} \right)