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)         \[\frac{7M{{L}^{2}}}{12}\]

E)  \[\frac{7M{{L}^{2}}}{48}\]

Correct Answer: E

Solution :

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