The moment of inertia of a disc about an axis passing through its centre and normal to its plane is I. The disc is now folded along a diameter such that the two halves are mutually perpendicular. Its moment of inertia about this diameter will now be-
A)I
B)\[I/\sqrt{2}\]
C)\[I/2\]
D)\[I/4\]
Correct Answer:
C
Solution :
\[I=\frac{1}{2}\,\,m{{R}^{2}}\] For a diameter; \[I'=\frac{1}{2}\]