ABC is a plane lamina of the shape of an equilateral triangle. D and E are mid points of AB and AC respectively, 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 l_a. If part ADE is removed, the moment of inertia of the remaining part about the same axis is $\frac{{\mathrm{Nl}}_{\mathrm{a}}}{16}$ where N is an integer. Value of N is ________

Divide the larger plate into four equal plates as shown. 

If  ${\mathrm{I}}_{\mathrm{a}}$ is moment of inertia of bigger plate, we may write
${\mathrm{I}}_{\mathrm{a}}=\mathrm{\beta m}{l}^2$
Where, β is a shape factor and $l$ is side length of the triangle. 

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

${\mathrm{I}}_{\mathrm{s}}=\mathrm{\beta }\frac{\mathrm{m}}{4}{\left(\frac{l}{2}\right)}^2=\frac{{\mathrm{I}}_{\mathrm{a}}}{16}$

So their contribution to ${\mathrm{I}}_{\mathrm{a}}$ can be written as

${\mathrm{I}}_{\mathrm{a}}=\frac{{\mathrm{I}}_{\mathrm{a}}}{16}+3\left[\left(\frac{{\mathrm{I}}_{\mathrm{a}}}{16}\right)+\mathrm{x}\right]$ [Using parallel axis theorem]

$\Rightarrow {\mathrm{I}}_{\mathrm{a}}-\frac{{\mathrm{I}}_{\mathrm{a}}}{16}-3\left(\frac{{\mathrm{I}}_{\mathrm{a}}}{16}\right)=3\mathrm{x}$

$\Rightarrow \mathrm{x}=\frac{{\mathrm{I}}_{\mathrm{a}}}{4}$

We require, 

${\mathrm{I}}_{\mathrm{remaining} \mathrm{plate} }={\mathrm{I}}_{\mathrm{a}}-{\mathrm{I}}_{\mathrm{small} \mathrm{plate}}$

$={\mathrm{I}}_{\mathrm{a}}-\left\{\frac{{\mathrm{I}}_{\mathrm{a}}}{16}+\mathrm{x}\right\}={\mathrm{I}}_{\mathrm{a}}-\left\{\frac{{\mathrm{I}}_{\mathrm{a}}}{16}+\frac{{\mathrm{I}}_{\mathrm{a}}}{4}\right\}=\frac{11{\mathrm{I}}_{\mathrm{a}}}{16}$