A) \[\frac{7M{{L}^{2}}}{48}\]
B) \[\frac{M{{L}^{2}}}{9}\]
C) \[\frac{M{{L}^{2}}}{12}\]
D) \[\frac{M{{L}^{2}}}{3}\]
Correct Answer: B
Solution :
According to the theorem of parallel axis \[{{I}_{AB}}={{I}_{G}}+M{{a}^{2}}\] where, I = moment of inertia of rod about its centre of gravity \[\therefore \] \[{{I}_{AB}}=\frac{M{{L}^{2}}}{12}+M{{\left( \frac{L}{2}-\frac{L}{3} \right)}^{2}}\] \[=\frac{M{{L}^{2}}}{12}+\frac{M{{L}^{2}}}{36}=\frac{M{{L}^{2}}}{9}\]You need to login to perform this action.
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