Physics - Elasticity Question with Solution | TestHub

Consider a small section$dx$of the bar at a distance$x$from B. The weight of the bar for a length$x$is,$W=\left(\frac{mg}{L}\right)x.$

Elongation in$x$will be

$de=\left(\frac{W}{AY}\right)dx=\left(\frac{mg}{LAY}\right)dx\cdot x$.

Total elongation of the bar can be obtained by integrating this expression for$x=0 to x=L.$

$\therefore \mathrm{\Delta }l= \underset{x=0}{\overset{x=L}{\int }}de=\left(\frac{mg}{LAY}\right) \underset{0}{\overset{L}{\int }}x\cdot dx$

$\Rightarrow \mathrm{\Delta }l=\frac{mgL}{2AY}$