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The moment of inertia of a uniform rectangular plate about an axis passing through its midpoint and parallel to its length l is

\[\begin {array} {1 1} (a)\;\frac{Mb^2}{4} & \quad (b)\;\frac{Mb^2}{6} \\ (c)\;\frac{Mb^2}{12} & \quad  (d)\;\frac{Ml^2}{12} \end {array}\]

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Divide plate into strips of man dm.
Parallel to the breadth
$dI= \large\frac{dmb^2}{12}$
$I= \int \large\frac{dm b^2}{12}$
$\quad= \large\frac{Mb^2}{12}$
answered Nov 26, 2013 by meena.p

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