#### ABC is a plane lamina of the shape of an equilateral triangle. D, E are midpoints of AB, AC and G is the centroid of the lamina. Moment of inertia of lamina about an axis passing through G and perpendicular to the plane ABC $I_{0}$. If the part ADE is removed, the moment of inertia of the remaining part about the same axis is $\frac{NI_{0}}{16}$ where N is an integer. Value of N is _____________ Option: 1 11 Option: 2 9 Option: 3 8 Option: 4 7

Let the side of the triangle is a and mass is m

\begin{aligned} &\text { MOI of plate ABC about centroid }\\ &\mathrm{I}_{0}=\frac{\mathrm{m}}{3}\left(\left(\frac{\mathrm{a}}{2 \sqrt{3}}\right)^{2} \times 3\right)=\frac{\mathrm{ma}^{2}}{12} \end{aligned}

Triangle ADE is also an equilateral triangle of side  $\frac{a}{2}$.

Let the moment of inertia of triangular plate ADE about its centroid (G') is $I_1$ and mass is  $m_1$

$\dpi{150} \begin{array}{l} \mathrm{m}_{1}=\frac{\mathrm{m}}{\frac{\sqrt{3} \mathrm{a}^{2}}{4}} \times \frac{\sqrt{3}}{4}\left(\frac{\mathrm{a}}{2}\right)^{2}=\frac{\mathrm{m}}{4} \\ \\ \mathrm{I}_{1}=\frac{\mathrm{m}_{1}}{12}\left(\frac{\mathrm{a}}{2}\right)^{2}=\frac{\mathrm{m}}{4 \times 12} \frac{\mathrm{a}^{2}}{4}=\frac{\mathrm{ma}^{2}}{192} \end{array}$

$\dpi{120} \text {from figure distance } \mathrm{GG}^{\prime}=\frac{\mathrm{a}}{\sqrt{3}}-\frac{\mathrm{a}}{2 \sqrt{3}}=\frac{\mathrm{a}}{2 \sqrt{3}}$

\begin{aligned} &\text { so MOI of part ADE about centroid G is }\\ &\mathrm{I}_{2}=\mathrm{I}_{1}+\mathrm{m}_{1}\left(\frac{\mathrm{a}}{2 \sqrt{3}}\right)^{2}=\frac{\mathrm{ma}^{2}}{192}+\frac{\mathrm{m}}{4} \cdot \frac{\mathrm{a}^{2}}{12}=\frac{5 \mathrm{ma}^{2}}{192} \end{aligned}

now MOI of remaining part

\begin{aligned} &I_{re}=\frac{m a^{2}}{12}-\frac{5 m a^{2}}{192}=\frac{11 m a^{2}}{12 \times 16}=\frac{11 I_{0}}{16} \\ \Rightarrow & N=11 \end{aligned}