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A circular hole of radius \frac{R}{4} is made in a thin uniform disc having mass M and radius R, as shown in figure.  The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is :

  • Option 1)

    \frac{219 MR^{2}}{256}

  • Option 2)


  • Option 3)


  • Option 4)



Answers (1)


As we learnt in

Paraller Axis Theorem -

I_{b\: b'}=I_{a\: a'}+mR^{2}

- wherein

b\: b' is axis parallel to a\: a' & a\: a' an axis passing through centre of mass.



 Moment of inertia of complete disc about O point.

\sigma_{1}=\frac{m}{A}=\frac{M}{\pi R^{2}}

\sigma_{2}=\frac{m}{A'}=\frac{M'}{\frac{\pi R^{2}}{16}}

\sigma_{2}=\frac{16M'}{\pi R^{2}}

\sigma_{1}=\sigma_{2}=\frac{M}{\pi R^{2}}=\frac{16M'}{\pi R^{2}}


M.O.I. of removed disc about its axis

I=\frac{1}{2}\frac{M}{16}\times\left(\frac{R}{4} \right )^{2}=\frac{MR^{2}}{512}

M.O.I. of removed disc about O


Where x=\frac{3R}{4}, I_{removed}=\frac{MR^{2}}{16}+\frac{M}{16}\left(\frac{3R}{4} \right )^{2}=\frac{19MR^{2}}{512}

\therefore\ \; I_{rem}=\frac{1}{2}MR^{2}-\frac{19MR^{2}}{512}=\frac{237MR^{2}}{512}

Option 1)

\frac{219 MR^{2}}{256}

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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