A) \[\frac{1}{2}M{{L}^{2}}\]
B) \[\frac{1}{3}M{{L}^{2}}\]
C) \[\frac{2}{3}M{{L}^{2}}\]
D) \[\frac{1}{4}M{{L}^{2}}\]
Correct Answer: A
Solution :
Moment of inertia of rod BC about an axis perpendicular to plane of triangle ABC and passing through the mid-point of rod BC (i.e., D) is \[{{l}_{1}}=\frac{{{m}^{2}}}{12}\] From theorem of parallel axes, moment of inertia of this rod about the asked axis is \[{{l}_{2}}={{l}_{1}}+M{{r}^{2}}=\frac{{{M}^{2}}}{12}+M\left( \frac{{}}{2\sqrt{3}} \right)=\frac{{{M}^{2}}}{6}\] \[\therefore \] Moment of inertia of all three rod is \[l=3{{l}_{2}}=3\left( \frac{{{M}^{2}}}{6} \right)=\frac{{{M}^{2}}}{2}\]You need to login to perform this action.
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