A) \[\frac{\left( M+3m \right){{L}^{2}}}{12}\]
B) \[\frac{\left( M+6m \right){{L}^{2}}}{12}\]
C) \[\frac{M{{L}^{2}}}{4}\]
D) \[\frac{M{{L}^{2}}}{12}\]
Correct Answer: B
Solution :
Moment of inertia of rod about XY \[{{I}_{1}}=\frac{M{{L}^{2}}}{12}\] Moment of intertia of two masses \[{{I}_{2}}=m{{\left( \frac{L}{2} \right)}^{2}}+m{{\left( \frac{L}{2} \right)}^{2}}=\frac{m{{L}^{2}}}{2}\] \[I={{I}_{1}}+{{I}_{2}}=\frac{M{{L}^{2}}}{12}+\frac{m{{L}^{2}}}{2}\] \[=\frac{(M+6m){{L}^{2}}}{12}\]
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