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A uniform steel rod of length 1 m and area of cross section $20 c m^{2}$ is hanging from a fixed support. Find the increase in the length of the rod $d u e t o i t s o w n w e i g h t$ $(Y_{s t e e l} = 2.0 \times 10^{11} N m^{- 2}, ρ_{s t e e l} = 7.85 \times 10^{3} k g m^{- 3})$

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$1.923 \times 10^{- 5} c m$

$1.123 \times 10^{- 5} c m$

$3.123 \times 10^{- 5} c m$

$2.923 \times 10^{- 5} c m$

answer is A.

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Detailed Solution

$e = \frac{L^{2} ρ g}{2 Y} e = \frac{1^{2} (7.85 \times 10^{3}) 10}{2 (2 \times 10^{11})} = 1.923 \times 10^{- 7} m = 1.923 \times 10^{- 5} c m$

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