question_answer

The moment of inertia of a thin rod of mass M and length L, about an axis perpendicular to  the rod at a distance \[\frac{L}{4}\] from one end is

A)  \[\frac{M{{L}^{2}}}{6}\]               

B)  \[\frac{M{{L}^{2}}}{12}\]

C)  \[\frac{7M{{L}^{2}}}{24}\]                           

D)  \[\frac{7M{{L}^{2}}}{48}\]

Correct Answer: D

Solution :

                 \[{{I}_{CD}}={{I}_{cm}}+M{{\left( \frac{L}{4} \right)}^{2}}\] \[=\frac{M{{L}^{2}}}{12}+\frac{M{{L}^{2}}}{16}=\frac{7M{{L}^{2}}}{48}\]