A uniform heavy rod of mass 20kg . Cross sectional area 0 . 4 m ^ 2...

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)$then, using Hooke's law,</p><p>$Y = \dfrac{mgx d x}{Al d \left(∆ l\right)} ⇒ d \left(∆ l\right) = \dfrac{max d x}{AlY} ⇒ {∫}_{0}^{∆ l} d \left(∆ l\right) = {∫}_{0}^{l}\dfrac{mgx d x}{AlY} ⇒ ∆ l = \dfrac{mgl}{2AY}$</p><p>$⇒ \Deltal = \dfrac{20 \times 10 \times 20}{2 \times 0 . 4 \times 2 \times {10}^{11}}$</p><p>$⇒ \Deltal = 25 \times {10}^{- 9}$</p><p>$⇒ x = 25