The moment of inertia of a circular ring about an axis passing through its centre and normal to its plane is 200 gm cm^{2}, then its moment of inertia about a diameter is 
 

  • Option 1)

    400 gm cm^{2}

  • Option 2)

    300 gm cm^{2}

  • Option 3)

    200 gm cm^{2}

  • Option 4)

    100 gm cm^{2}

 

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.

 I_{z}=I_{x}+I_{y}

200=I_{D}+I_{D}

200=2I_{D}\:\:\:\:\:\Rightarrow I_{D}=100gm\:cm^{2}

 


Option 1)

400 gm cm^{2}

This option is incorrect.

Option 2)

300 gm cm^{2}

This option is incorrect.

Option 3)

200 gm cm^{2}

This option is incorrect.

Option 4)

100 gm cm^{2}

This option is correct.

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