JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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
- A \(\frac{{219\,M{R^2}}}{{256}}\)
- B \(\frac{{237\,M{R^2}}}{{512}}\)
- C \(\frac{{19\,M{R^2}}}{{512}}\)
- D \(\frac{{197\,M{R^2}}}{{256}}\)
Answer & Solution
Correct Answer
(B) \(\frac{{237\,M{R^2}}}{{512}}\)
Step-by-step Solution
Detailed explanation
Moment of Inertia of complete disc about \('O'\) point \({I_{total}} = \frac{{M{R^2}}}{2}\) Radius of removed disc \(= R/4\) Mass of removed disc \(= M/16\) [As \(M\) \( \propto \) \({{R^2}}\)] \(M.I.\) of removed disc about its own axis \((O')\)…
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