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ABC is a plane lamina of the shape of an equilateral triangle DE are midpoints of AB and 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 same axis is $\frac{N I_{0}}{16}$ where N is an integer value of N is____

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answer is 11.

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Detailed Solution

Moment of inertia of lamina about an axis passing through centroid “G”

$I_{0} = K m l^{2}$

Moment of inertia of DEF

$I_{1} = (\frac{m}{4}) {(\frac{l}{2})}^{2}$

$I_{1} = \frac{I_{0}}{16}$

Let $I_{A D E} = I_{B D F} = I_{E F G} = I_{2}$

$∴ 3 I_{2} + I_{1} = I_{0}$

$∴ 3 I_{2} + \frac{I_{0}}{16} = I_{0}$

$I_{2} = \frac{5 I_{0}}{16}$

After removal of part ADE

$I = 2 I_{2} + I_{1}$

$= 2 [\frac{5 I_{0}}{16}] + \frac{I_{0}}{16}$

$I = \frac{11 I_{0}}{16}$

Given $I = \frac{N I_{0}}{16}$

N = 11

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