question_answer

The moment of inertia of uniform thin rod of length L and mass M about an axis passing through a point at a distance \[\frac{L}{3}\] from one of its ends and perpendicular to the rod is

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                \[\therefore \]  \[{{I}_{AB}}=\frac{M{{L}^{2}}}{12}+M{{\left( \frac{L}{2}-\frac{L}{3} \right)}^{2}}\]                 \[=\frac{M{{L}^{2}}}{12}+\frac{M{{L}^{2}}}{36}=\frac{M{{L}^{2}}}{9}\]