• # question_answer A thin disc of mass 9M and radius R from which a disc of radius $R/3$ is cut, as shown in figure. Then moment of inertia of the remaining disc about 0, perpendicular to the plane of disc is- A) $4\text{ }M{{R}^{2}}$                      B) $9\text{ }M{{R}^{2}}$C) $\frac{37}{9}\,M{{R}^{2}}$              D) $\frac{40}{9}M{{R}^{2}}$

$I={{I}_{1}}-{{I}_{2}}$ $=\,\,4\,M{{R}^{2}}$