An equilateral triangle ABC is cut from a thin solid sheet of wood (see figure). D, E and F are the mid-points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of triangle is $I_{0}$ . If the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. Then

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$I = \frac{9}{16} I_{0}$

$I = \frac{3}{4} I_{0}$

$I = \frac{15}{16} I_{0}$

$I = \frac{I_{0}}{4}$

answer is D.

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Detailed Solution

Suppose M is a is side of larger triangle, then $\frac{M}{4}$ and $\frac{a}{2}$ will be mass and side length of smaller triangle.
$\frac{I_{r e m o v e d}}{I_{o r i g i n a l}} \frac{\frac{M}{4}}{M} . \frac{{(\frac{a}{2})}^{2}}{{(a)}^{2}}$
$I_{r e m o v e d} = \frac{I_{0}}{16}$
So, $I = I_{0} - \frac{I_{0}}{16} = \frac{15 I_{0}}{16}$