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 $\frac{N{I}_0}{16}$ where N is an integer. Value of N is __________ .

Divide the larger plate into four equal plates as shown.
If  $I_0$ is moment of inertia of bigger plate, we may write
$I_0=\beta m{a}^2$ 
Where  $\beta$ is a shape factor. 

So each small plate will have moment of inertia about its own centroid as

  $=I=\beta \frac{m}{4}{\left(\frac{a}{2}\right)}^2=\frac{I_0}{16}$.
So their contribution to $I_0$  can be written as

  $\begin{array}{l}{I}_0=\frac{I_0}{16}+3\left[\left(\frac{I_0}{16}\right)+x\right] \left[U\sin g parallel axis theorem\right]\text{}\\ I_0-\frac{I_0}{16}-3\left(\frac{I_0}{16}\right)=3x\end{array}$
 $\Rightarrow x=\frac{I_0}{4}$
We require  $I_{remaining plate}=I_0-I_{small plate}$
 $=I_0-\left\{\frac{I_0}{16}+x\right\}=I_0-\left\{\frac{I_0}{16}+\frac{I_0}{4}\right\}=\frac{11I_0}{16}$