Question

A uniform heavy rod of mass $20kg$. Cross sectional area $0.4m^2$ 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 $Y=2\times 10^{11}N{m}^{-2}$ and $g=10m{s}^{-2}$ )

Answer 1

Answer: 25

$Y=\frac{T}{A}\frac{dx}{dy}$
$m=20kg$
$A=0.4m^2$
$1=20m$
let extension is dy in length $dx$
$Y=\frac{\text{ stress }}{\text{ strain }}$
$Y=\frac{\frac{T}{A}}{\frac{dy}{dx}}=\frac{T}{A}\cdot \frac{dx}{dy}$
$dy=\frac{Tdx}{AY}$
Tension at a distance $x$ from lower end
$=\frac{mg}{\ell }x$
So. $\underset{0}{\overset{Nl}{\int }}dy=\underset{0}{\overset{\ell }{\int }}\frac{mg}{\ell }x\frac{dx}{AY}$
$\Delta \ell =\frac{mg}{\ell AY}{\left[\frac{x^2}{2}\right]}_0^{\ell }$
$\Delta \ell =\frac{mg\ell }{2AY}$
$\Delta \ell =\frac{20\times 10\times 20}{2\times 0.4\times 2\times 10^{11}}$
$2500\times 10^{-11}$
$\Delta \ell =25\times 10^{-9}$
$=x\times 10^{-9}$
$x=25$