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A uniform heavy rod of weight 10 kg ms^–2, cross-sectional area 100 cm² 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

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5 × 10^–8 m

5 × 10^–10 m

4 × 10^–8 m

2 × 10^–9 m

answer is B.

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Detailed Solution

We know that,

$∆ l = \frac{WL}{2 AY} = \frac{10 \times 1}{2 \times 5} \times 100 \times 10^{- 4} \times 2 \times 10^{11} = 5 \times 10^{- 10} m$

Hence the correct answer is $5 \times 10^{- 10} m .$

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