A) \[4M{{R}^{2}}\]
B) \[\frac{40}{9}M{{R}^{2}}\]
C) \[10M{{R}^{2}}\]
D) \[\frac{37}{9}M{{R}^{2}}\]
Correct Answer: D
Solution :
\[\frac{{{m}_{1}}}{{{A}_{1}}}=\frac{{{m}_{2}}}{{{A}_{2}}}\] \[\frac{9M}{\pi {{R}^{2}}}=\frac{m}{\pi {{\left( \frac{R}{2} \right)}^{2}}}\] \[{{\operatorname{m}}^{2}}=M\] \[{{\operatorname{I}}_{rem}}={{I}_{wide}}-{{I}_{removed}}\] \[=\frac{1}{2}9M{{R}^{2}}-\left[ \frac{1}{2}M{{\left( \frac{R}{3} \right)}^{2}}+\frac{1}{2}M{{\left( \frac{2R}{3} \right)}^{2}} \right]\] \[{{\operatorname{I}}_{rem}}=4M{{R}^{2}}\]You need to login to perform this action.
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