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Moment of inertia of a uniform rod of length L and mass M, about an axis passing through $L/4$ from one end and perpendicular to its length is

\[\begin {array} {1 1} (a)\;\frac{7}{36}\;ML^2 & \quad (b)\;\frac{7}{48}\;ML^2 \\ (c)\;\frac{11}{48}\;ML^2 & \quad  (d)\;\frac{Ml^2}{12} \end {array}\]

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$I= I_{cm}+M (\frac{l}{4})^2$
$\quad= \large\frac{Ml^2}{12}+\frac{Ml^2}{16}$
$\quad =\large\frac{4Ml^2+3Ml^2}{48}$
$\quad= \large\frac{7ml^2}{48}$
answered Nov 26, 2013 by meena.p

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