A) \[\frac{40}{9}M{{R}^{2}}\]
B) \[M{{R}^{2}}\]
C) \[4M{{R}^{2}}\]
D) \[\frac{4}{9}M{{R}^{2}}\]
Correct Answer: A
Solution :
The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre \[I={{I}_{1}}-{{I}_{2}}\] \[=\frac{9M{{R}^{2}}}{2}-\frac{M{{R}^{2}}}{18}\] \[=\frac{81M{{R}^{2}}-M{{R}^{2}}}{18}\] \[=\frac{40M{{R}^{2}}}{9}\]
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