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A thin wire of length l and mass m is bent in the form of a semicircle as shown. Its moment of inertia about an axis joining its free ends will be

\[\begin {array} {1 1} (a)\;Ml^2 & \quad (b)\;zero \\ (c)\;\frac{ml^2}{\pi^2} & \quad  (d)\;\frac{ml^2}{2\pi ^2} \end {array}\]

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$l=\pi r$
$r= \large\frac{l}{\pi}$
Symmetry M.I of full ring about $YY'$
$\quad=(2m) \large\frac{r^2}{2}$
M.I of half ring $= m \bigg(\large\frac{r^2}{2}\bigg)$
$\qquad= m \bigg(\large\frac{l^2}{2 \pi ^2}\bigg)$
answered Nov 26, 2013 by meena.p
edited Jun 19, 2014 by lmohan717
 

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