A) \[\frac{2}{3}M\ell \]
B) \[\frac{M\ell }{4}\]
C) \[\frac{M{{\ell }^{2}}}{2}\]
D) None of these
Correct Answer: C
Solution :
\[I={{I}_{1}}+{{I}_{2}}+{{I}_{3}}\] \[=\frac{M{{(\ell \,\sin \,60{}^\circ )}^{2}}}{3}+\frac{M{{(\ell \,\sin 60{}^\circ )}^{2}}}{3}+0=\frac{M{{\ell }^{2}}}{2}\]You need to login to perform this action.
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