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Moment of inertia of a square lamina about one of the diagonals is :

\[\begin {array} {1 1} (a)\;\frac{Ml^2}{6} & \quad (b)\;\frac{Ml^2}{12} \\ (c)\;\frac{Ml^2}{3} & \quad  (d)\;\frac{Ml^2}{24} \end {array}\]

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Using perpendicular Axes theorems
$Iz=I_{xx'}+I_{yy'}$
$\large\frac{Ml^2}{6}$$=2 I xx^{1} \qquad \bigg[I_{xx'}=I_{yy'} \;by\;symmetry\bigg]$
$\therefore I_{xx^1}=\large\frac{Ml^2}{12}$

 

answered Nov 26, 2013 by meena.p
 

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