Question 49 of 118

A uniform heavy rod of mass $20kg$. Cross sectional area $0.4{m}^{2}$ and length $20m$ is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is $x\times {10}^{-9}m$. The value of $x$ is _____ .
(Given. Young's modulus $Y=2\times {10}^{11}{Nm}^{-2}$ and $g=10m{s}^{-2}$ )

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Answer: 25

Tension at a distance $x$ from lower end $=\dfrac{mg}{l}x$
If the elongation in the element is taken as $d\left(∆l\right)$ then, using Hooke's law,
$Y=\dfrac{mgxdx}{Ald\left(∆l\right)} \Rightarrow d\left(∆l\right)=\dfrac{maxdx}{AlY} \Rightarrow {\int }_{0}^{∆l}d\left(∆l\right)={\int }_{0}^{l}\dfrac{mgxdx}{AlY} \Rightarrow ∆l=\dfrac{mgl}{2AY}$
$\Rightarrow \Delta l=\dfrac{20\times 10\times 20}{2\times 0.4\times 2\times {10}^{11}}$
$\Rightarrow \Delta l=25\times {10}^{-9}$
$\Rightarrow x=25$