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A rod of mass M of radius r,length l made up of material of young modulus Y is rotating in horizontal plane with angular velocity $\omega$.

$\begin{array}{1 1}(A)\;\large\frac{M\omega^2l^2}{3\pi r^2}\\(B)\;\large\frac{M\omega^2l^2}{2\pi r^2}\\(C)\;\large\frac{M\omega^2r^2}{3\pi l^2}\\(D)\;\large\frac{M\omega^2r}{2\pi l}\end{array} $

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Considering an element of $dx$ length.
Force on it =$\int\limits_x^l dm\omega^2Y$
$\Rightarrow \int_x^l \large\frac{M}{l}$$dy\omega^2y$
$\Rightarrow [\large\frac{M\omega^2}{l}\frac{Y^2}{2}]_x^l$
$\Rightarrow \large\frac{M\omega^2}{2l}$$(l^2-x^2)$
Let the extension in element $dx$ be $d\alpha$
As $Y=(\large\frac{f}{a})\frac{l}{\Delta l}$
$\Delta l=(\large\frac{f}{a})\frac{l}{Y}$
For element of $dx$ length
$d\alpha=\large\frac{M\omega^2(l^2-x^2)dx}{2laY}$
$\alpha=\int\limits_0^l \int \large\frac{M\omega^2}{2la}$$(l^2-x^2)dx$
$\;\;\;=\large\frac{M\omega^2}{2la}[$$l^3-\large\frac{l^3}{3}]$
$\;\;\;=\large\frac{M\omega^2l^2}{3a}$
As $a=\pi r^2$
$\;\;\;=\large\frac{M\omega^2l^2}{3\pi r^2}$
Hence (A) is the correct answer.
answered May 9, 2014 by sreemathi.v
 

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