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)

    \frac{7}{8}I{\circ} 

  • Option 2)

  • Option 3)

  • Option 4)

 

Answers (1)

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)

\frac{7}{8}I{\circ} 

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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