question_answer

The moment of inertia of a rod (length I mass m) about an axis perpendicular to the length of the rod and passing through a J point equidistant from its mid-point and one end, is

A)  \[\frac{m{{l}^{2}}}{12}\]                                              

B)  \[\frac{7}{48}m{{l}^{2}}\]

C)  \[\frac{13}{48}m{{l}^{2}}\]                         

D)  \[\frac{19}{48}m{{l}^{2}}\]

E)  \[\frac{14}{48}m{{l}^{2}}\]

Correct Answer: B

Solution :

                Moment of inertia of thin rod \[I=\frac{m{{l}^{2}}}{12}\] Distance of axis CD from AB\[=\frac{L}{4}\] From theorem of parallel axis, we have \[{{I}_{CD}}={{I}_{AB}}+M{{\left( \frac{l}{4} \right)}^{2}}\]        \[=\frac{m{{l}^{2}}}{12}+\frac{m{{l}^{2}}}{16}\]