question_answer

The radius of gyration of a rod of length \[L\] and mass \[M\] about an axis perpendicular to its length and passing through a point at a distance \[L/3\] from one of its ends is

A) \[\frac{\sqrt{7}}{6}L\]                   

B) \[\frac{{{L}^{2}}}{9}\]

C) \[\frac{L}{3}\]                                   

D) \[\frac{\sqrt{5}}{2}L\]

Correct Answer: C

Solution :

Moment of inertia of the rod about a perpendicular axis \[PQ\] passing through centre of mass\[C\] Let \[N\] be the point which divides the length of rod \[AB\] in ratio\[1:3\]. This point will be at a distance \[\frac{L}{6}\] from\[C\]. Thus, the moment of inertia about an axis parallel to \[PQ\] and passing through the point\[N\].                 \[I={{I}_{CM}}+M{{\left( \frac{L}{6} \right)}^{2}}\]                    \[=\frac{M{{L}^{2}}}{12}+\frac{M{{L}^{2}}}{36}=\frac{M{{L}^{2}}}{9}\] If \[K\] be the radius of gyration, then                 \[K=\sqrt{\frac{I}{M}}=\sqrt{\frac{{{L}^{2}}}{9}}=\frac{L}{3}\]