JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2025
- Medium
- MCQ
A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius \(\mathrm{R} / 2\) is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.
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Answer & Solution
Correct Answer
(D) \(\frac{13}{32} \mathrm{MR}^2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & I_{\text {complete }}=\frac{M R^2}{2} \\ & \begin{aligned} I_{\text {removed }} & =\left(\frac{M}{4}\right)\left(\frac{R}{2}\right)^2 \frac{1}{2}+\left(\frac{M}{4}\right)\left(\frac{R}{2}\right)^2 \\ & =\frac{3 M R^2}{32} \\ I=\frac{M R^2}{2} & -\frac{3 M R^2}{32}=\frac{13}{32} M R^2\end{aligned}\end{aligned}\)
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