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23/01/2022 2:11 pm
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A uniform heavy rod of weight 10 kg ms-2, cross-sectional area 100 cm2 and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2 × 1011 Nm-2. Neglecting the lateral contraction, find the elongation of rod due to its own weight.
(1) 2 × 10–9 m
(2) 5 × 10–8 m
(3) 4 × 10–8 m
(4) 5 × 10–10 m
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23/01/2022 2:18 pm
Correct answer: (4) 5 × 10–10 m
Explanation:
We know that,
Δℓ = \(\frac{WL}{2AY}\)
Δℓ = \(\frac{10 \times 1}{2 \times 5}\) x 100 x 10–4 x 2 x 1011
Δℓ = \(\frac{1}{2}\) x 10–9
= 5 x 10–10 m