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  • From a circular ring  of mass 'M' and radius 'R' an are corresponding to a 900 sector is removed. The moment of inertia of the remaining part of the ring about an ax

From a circular ring  of mass 'M' and radius 'R' an are corresponding to a 900 sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the center of the ring and perpendicular to the plane of the ring is 'K'times MR^{2},Then the value of 'K' is :  

Option: 1

\frac{3}{4}


Option: 2

\frac{7}{8}


Option: 3

\frac{1}{4}


Option: 4

\frac{1}{8}


Answers (1)

best_answer

Remaining mass={M}'=\frac{3}{4}M

I={M}'R^{2}=\frac{3}{4}MR^{2}-K\: MR^{2}

K=\frac{3}{4}

Posted by

himanshu.meshram

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