A) \[\frac{17}{15}M{{R}^{2}}\]
B) \[\frac{137}{15}M{{R}^{2}}\]
C) \[\frac{209}{15}M{{R}^{2}}\]
D) \[\frac{152}{15}M{{R}^{2}}\]
Correct Answer: A
Solution :
\[I=2{{I}_{sp}}+{{I}_{rod}}\] \[=2\left[ \frac{2}{5}M{{R}^{2}}+M{{(2R)}^{2}} \right]+\frac{M{{(2R)}^{2}}}{12}\] \[=\,\,\,2\left[ \frac{22}{5}M{{R}^{2}} \right]+\frac{M{{R}^{2}}}{3}\] \[=\,\,\,\frac{(132+5)}{15}M{{R}^{2}}\] \[I=\frac{137}{15}M{{R}^{2}}\]You need to login to perform this action.
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