A) \[\frac{2}{3}M{{l}^{2}}\]
B) \[\frac{13}{3}M{{l}^{2}}\]
C) \[\frac{1}{3}M{{l}^{2}}\]
D) \[\frac{4}{3}M{{l}^{2}}\]
Correct Answer: D
Solution :
[d] Moment of inertia of a thin rod of length \[l\] about an axis passing through centre and perpendicular to the rod \[=\frac{1}{12}M{{l}^{2}}\] Thus moment of inertia of the frame. \[\frac{m{{l}^{2}}}{12}+\frac{m{{l}^{2}}}{4}=\frac{4m{{l}^{2}}}{12}=\frac{m{{l}^{2}}}{3}\] Total M.I. \[=4\times \frac{m{{l}^{2}}}{3}\]
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