One quarter section is cut from a uniform circular disc of radius R. This section has a mass M. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is
- $\frac{1}{2}MR^2$
- $\frac{1}{4}MR^2$
- $\frac{1}{8}MR^2$
- $\sqrt 2 MR^2 $
The Correct Option is A
Solution and Explanation
Moment of inertia of the disc about the given axis
$ \, \, \, \, \, \, \, \, \, = \frac{1}{2} (4M) R^2 = 2MR^2$
$ \therefore $ Moment of inertia of quarter section of the disc
$ \, \, \, \, \, \, \, \, \, = \frac{1}{4} (2M R^2 )= \frac{1}{2}MR^2$
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