JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2018
- Hard
- MCQ
A thin circular disc is in the \(xy\) plane as shown in the figure. The ratio of its moment of inertia about \(z\) and \(z'\) axes will be

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Answer & Solution
Correct Answer
(C) \(1:3\)
Step-by-step Solution
Detailed explanation
As we know, moment of inertia of a disc about an axis passing through \(C.G.\) and perpendicular to its plane, \({I_z}\, = \frac{{m{R^2}}}{2}\) Moment of inertia of a disc about a tangential axis perpendicular to its own plane,
\({I_z}'\, = \frac{3}{{20}}m{R^2}\)
\({I_z}/{I_z}'\, = \frac{{m{R^2}}}{2}/\frac{{3m{R^2}}}{2}\, = 1/3\) 
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