A) 4 MR2
B) \[\frac{40}{9}\]MR2
C) \[10M{{R}^{2}}\]
D) \[\frac{37}{9}M{{R}^{2}}\]
Correct Answer: A
Solution :
As the mass is uniformly distributed on the disc, so mass density (per unit area) \[=\frac{9M}{\pi {{R}^{2}}}\] Mass of removed portion\[=\frac{9M}{\pi {{R}^{2}}}\times \pi {{\left( \frac{R}{3} \right)}^{2}}\] So moment of inertia of the removed portion about the stated axis by theorem of parallel axis. \[{{I}_{1}}=\frac{1}{2}M{{\left( \frac{R}{3} \right)}^{2}}+M{{\left( \frac{2R}{3} \right)}^{2}}\] If the disc would not have been removed, then the moment of inertia of complete disc about the stated axis. \[{{I}_{2}}=\frac{1}{2}9M{{(R)}^{2}}\] So, the moment of inertia of the disc about required axis. \[I={{I}_{2}}-{{I}_{1}}\] \[=\frac{1}{2}9M{{(R)}^{2}}-\left[ \frac{1}{2}M{{\left( \frac{R}{3} \right)}^{2}}+\frac{1}{2}M{{\left( \frac{2R}{3} \right)}^{-2}} \right]\]\[I=4M{{R}^{2}}\]
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