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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 the triangle is I₀. If the smaller triangle DEF is removed form ABC, the moment of inertia of the remaining figure about the same axis is l then

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

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

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

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

answer is A.

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Detailed Solution

M.I of Triangle $= \frac{{Ma}^{2}}{12} = I_{o}$

M.I of removed Triangle $= \frac{1}{12} \frac{M}{4} {(\frac{a}{2})}^{2}$

$\frac{I_{removed}}{I_{original}} = \frac{1}{16}$

$I_{remaining} = I_{o} - \frac{I_{o}}{16} = \frac{15 I_{o}}{16}$

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