# Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane is Io as shown in the figure.  A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides.  Moment of inertia of the remaining part of lamina about the same axis is : Option 1)   Option 2) Option 3) Option 4)

As we learnt in

Perpendicular Axis theorem -

$I_{z}= I_{x}+I_{y}$

(for a body in XY plane )

- wherein

$I_{z}$ = moment of inertia about z axis

$I_{x}$ .$I_{y}$ :moment of inertia about x & y  axis in the plane of body respectively.

According to the theorem of perpendicular axes.  Moment of inertia of triangle (ABC)

$I_{0}=mR^{2}$

Moment of inertia of cavity (DEF)

$I_{DEF}=\frac{m}{4}\left ( \frac{1}{2} \right )^{2}=\frac{mR^{2}}{16}$

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

Therefore, Remaining part $=I_{remain}=I_{0}-\frac{I_{0}}{16}=\frac{15 I_{0}}{16}$

Option 1)

This is an incorrect option.

Option 2)

This is the correct option.

Option 3)

This is an incorrect option.

Option 4)

This is an incorrect option.

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