The radius of gyration of 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

Options

A. `(sqrt7)/(6)L`
B. `(L^(2))/(9)`
C. `(L)/(3)`
D. `(sqrt5)/(2)L`

Answer

C

Explanation

Moment of inertia of the rod about a perpendicular axis PQ passing through centre of mass C, `l_("CM")=(ML^(2))/(12)`
Let N be the point which divides the length of rod AB in ratio `1:3`.
This point will be a distance `(L)/(6)` from C.

Thus, the moment of inertia l' about an axis parallel to PQ and passing through the point N.
`l'=l_("CM")+M((L)/(6))^(2)=(ML^(2))/(12)+(ML^(2))/(36)=(ML^(2))/(9)`
If K be the radius of gyration, then `K=sqrt((l')/(M))=sqrt((L^(2))/(9))=(L)/(3)`