A) \[M{{R}^{2}}\]
B) \[4M{{R}^{2}}\]
C) \[\frac{4}{9}\,M{{R}^{2}}\]
D) \[\frac{40}{9}\,M{{R}^{2}}\]
Correct Answer: D
Solution :
\[\operatorname{Mass} of complete disc = 9M\] \[Mass of removed portion of disc =M\] Moment of inertia of complete disc at point O perpendicular to plane \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{l}_{1}}=\frac{1}{2}\,\left( 9M \right){{R}^{2}}\] \[=\,\,\,\frac{9}{2}\,M{{R}^{2}}\] Moment of inertia of the removed portion of disc about same axis \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{I}_{2}}=\frac{1}{2}M\,{{\left( \frac{R}{3} \right)}^{2}}\] \[=\,\,\,\frac{1}{18}\,M{{R}^{2}}\] Moment of inertia of remaining disc from same axis is \[I={{I}_{1}}-{{I}_{2}}=\frac{9}{2}M{{R}^{2}}-\frac{M{{R}^{2}}}{18}=\frac{40\,M{{R}^{2}}}{9}\]
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