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From a circular disc of radius R and mass 9 M , a small disc of mass M and radius (R)/(3) is removed concentrically . The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

Answer» <html><body><p>`(<a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a>)/(9) MR^(2)`<br/>`MR^(2)`<br/>` 4 MR^(2)`<br/>`(4)/(9) MR^(2)`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Mass of the <a href="https://interviewquestions.tuteehub.com/tag/disc-955290" style="font-weight:bold;" target="_blank" title="Click to know more about DISC">DISC</a> = 9 M <br/> Mass of removed portion of disc = M <br/> The moment of inertia of the complete disc about an axis passing through its centre O and <a href="https://interviewquestions.tuteehub.com/tag/perpendicular-598789" style="font-weight:bold;" target="_blank" title="Click to know more about PERPENDICULAR">PERPENDICULAR</a> to its plane is <br/> `I_(1) = (9)/(2) MR^(2)` <br/> Now, the moment of inertia of the disc with removed portion <br/> `I_(2) = (1)/(2) M ((R)/(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))^(2) = (1)/(18) MR^(2)` <br/> Therefore , moment of inertia of theremaining portion of disc about O is `I = I_(1) - I_(2) = (9 MR^(2))/(2) = (MR^(2))/(18) = (40 MR^(2))/(9)`</body></html>


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