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A uniform heavy rod of mass $20\,kg$. Cross sectional area $0.4\,m ^{2}$ and length $20\,m$ 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 $Y =2 \times 10^{11} Nm ^{-2}$ અને $\left.g=10\, ms ^{-2}\right)$
$28$
$25$
$24$
$23$
Solution
$Y =\frac{ T }{ A } \frac{ dx }{ dy }$
$m =20\,kg$
$A =0.4\,m^{2}$
$1=20\,m$
let extension is $dy$ in length $dx$
$Y =\frac{\text { stress }}{\text { strain }}$
$Y =\frac{\frac{ T }{ A }}{\frac{ d }{ dx }}=\frac{ T }{ A } \cdot \frac{ dx }{ dy }$
$dy =\frac{ Tdx }{ AY }$
Tension at a distance $x$ from lower end $=\frac{ mg }{\ell} x$
So. $\int_{0}^{\Delta l} dy =\int_{0}^{\ell} \frac{ mg }{\ell} x \frac{ dx }{ AY }$
$\Delta \ell=\frac{ mg }{\ell AY }\left[\frac{ x ^{2}}{2}\right]_{0}^{\ell}$
$\Delta \ell=\frac{ mg \ell}{2\,AY }$
$\Delta \ell=\frac{20 \times 10 \times 20}{2 \times 0.4 \times 2 \times 10^{11}}$
$2500 \times 10^{-11}$
$\Delta \ell=25 \times 10^{-9}$
$= x \times 10^{-9}$
