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Q)

Moment of inertia of a portion of a disc about an axis perpendicular to its plane; Mass of the object is M and radius R is

$\begin {array} {1 1} (a)\;\frac{MR^2}{2} & \quad (b)\;\frac{1}{6}\frac{MR^2}{2} \\ (c)\;6.\frac{MR^2}{2} & \quad (d)\;none\;of\;the \;above \end {array}$

$\therefore 6I= \large\frac{(6M)R^2}{2}$
$\therefore I= \large\frac{MR^2}{2}$