A uniform heavy rod of mass 20,kg. Cross sect | vedclass

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Question

$Y =\frac{ T }{ A } \frac{ dx }{ dy }$

$m =20\,kg$

$A =0.4\,m^{2}$

$1=20\,m$

let extension is $dy$ in length $dx$

$Y =\frac{	ext { str

Vedclass · Accepted Answer

$25$

Vedclass · Answer

$Y =\frac{ T }{ A } \frac{ dx }{ dy }$

$m =20\,kg$

$A =0.4\,m^{2}$

$1=20\,m$

let extension is $dy$ in length $dx$

$Y =\frac{	ext { stress }}{	ext { strain }}$

$Y =\frac{\frac{ T }{ A }}{\frac{ d }{ dx }}=\frac{ T }{ A } \cdot \frac{ dx }{ dy }$

$dy =\frac{ Tdx }{ AY }$

Tension at a distance $x$ from lower end $=\frac{ mg }{\ell} x$

So. $\int_{0}^{\Delta l} dy =\int_{0}^{\ell} \frac{ mg }{\ell} x \frac{ dx }{ AY }$

$\Delta \ell=\frac{ mg }{\ell AY }\left[\frac{ x ^{2}}{2}ight]_{0}^{\ell}$

$\Delta \ell=\frac{ mg \ell}{2\,AY }$

$\Delta \ell=\frac{20 	imes 10 	imes 20}{2 	imes 0.4 	imes 2 	imes 10^{11}}$

$2500 	imes 10^{-11}$

$\Delta \ell=25 	imes 10^{-9}$

$= x 	imes 10^{-9}$

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