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# A rod of radius 2mm,length 80cm density $5KKg/m^3$ and made up of material of $Y=1\times 10^{10}N/m^2$.Find the elongation due to its own weight.

$\begin{array}{1 1}(A)\;1.6\times 10^{-6}m\\(B)\;3.2\times 10^{-6}m\\(C)\;2.4\times 10^{-6}m\\(D)\;\text{Zero}\end{array}$

Considering an element of length dx at a distance $x$.Force on in it due to lower part is $s(l-x)ag$ .Let the elongation in this element be $\alpha$.
As $Y=(\large\frac{f}{a})\frac{l}{\Delta l}$
$Y=\large\frac{s(l-x)ag}{a}\frac{dx}{\alpha}$
$\alpha=\large\frac{Sg}{Y}$$(l-x)dx Total elongation is \int \alpha \int \alpha=\int\limits_0^l \large\frac{Sg}{Y}$$(l-x)dx$
$\Rightarrow \large\frac{Sg}{Y}$$[(lx-\large\frac{x^2}{2}]_0^l$
$\Rightarrow \large\frac{Sgl^2}{2Y}$
$\Rightarrow \large\frac{5\times 10^3\times 10\times (.8)^2}{2\times 1\times 10^{10}}$
$\Rightarrow \large\frac{5\times .64\times 10^{-6}}{2}$
$\Rightarrow 1.6\times 10^{-6}m$
Hence (A) is the correct answer.