Question

A uniform heavy rod of mass $20\text{kg}$, cross sectional area $0.4{\text{m}}^2$ and length $20\text{m}$ 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 $Y=2\times {10}^{11}{\text{Nm}}^{-2}$ and $g=10{\text{ms}}^{-2}$)

Answer 1

Using,

$\Delta L=\frac{FL}{AY}=\frac{T_{avg}L}{AY}=\frac{MgL}{2AY}$

$=\frac{20\times 10\times 20}{2\times 0.4\times 2\times {10}^{11}}=\frac{4\times {10}^3\times {10}^{-11}}{4\times 0.4}$

$=2.5\times {10}^{-8}=25\times {10}^{-9}$

$\therefore x=25$