Question 68 of 280

$ABC$ is a plane lamina of the shape of an equilateral triangle. $D,E$ are mid-points of $AB,AC$ and $G$ is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through $G$ and perpendicular to the plane $ABC$ is ${I}_{0}$. If part $ADE$ is removed, the moment of inertia of the remaining part about the same axis is $\dfrac{N{I}_{0}}{16}$ where $N$ is an integer. Value of $N$ is:

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Answer: 11

Let mass of lamina $=m$, and length of side $=l$, then moment of inertila of lamina about an axis passing through $G$ and perpendicular to the plane.
${I}_{0}\propto m{l}^{2}$
${I}_{0}=km{l}^{2}$

Let moment of inertila of $DEF={I}_{1}$ about $G$
then, ${I}_{1}=\dfrac{{I}_{0}}{16}$
then,${I}_{ADE}={I}_{BDF}={I}_{EFC}={I}_{2}$
$3{I}_{2}+{I}_{1}={I}_{0}$
$3{I}_{2}+\dfrac{{I}_{0}}{16}={I}_{0}$
$\Rightarrow$ ${I}_{2}=\dfrac{5{I}_{0}}{16}$
So, Moment of inertila of $DECB=2{I}_{2}+{I}_{1}2\left(\dfrac{5{I}_{0}}{16}\right)+\left(\dfrac{{I}_{0}}{16}\right)$
$=\dfrac{11{I}_{0}}{16}$