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From a uniform circular disc of radius R and mass 9 M, a small disc of radius \frac{R}{3} is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is :

  • Option 1)

    \frac{37}{9} MR^{2}

  • Option 2)

    4 MR2

  • Option 3)

    \frac{40}{9} MR^{2}

  • Option 4)

    10 MR2

 

Answers (2)

best_answer

As we learnt that

 

Moment of inertia for disc -

I=\frac{1}{2} MR^{2}

 

- wherein

About an axis perpendicular to the plane of disc & passing through its centre .

 

 

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.

 

 I=\frac{9}{2}MR^{2}-[\frac{M(\frac{R}{3})^{2}}{2}+M(\frac{2R}{3})^{2}]

    =MR^{2}[\frac{9}{2}-\frac{1}{18}-\frac{4}{9}]

I=4MR^{2}

 

 

 


Option 1)

\frac{37}{9} MR^{2}

This is incorrect

Option 2)

4 MR2

This is correct

Option 3)

\frac{40}{9} MR^{2}

This is incorrect

Option 4)

10 MR2

This is incorrect

Posted by

Plabita

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