Let M be the mass and L be the length of a thin uniform rod. In first case, axis of rotation is passing through centre and perpendicular to the length of the rod. In second case, axis of rotation is passing through one end and perpendicular to the length of the rod. The ratio of radius of gyration in first case to second case is

Options

A. 1
B. `(1)/(2)`
C. `(1)/(4)`
D. `(1)/(8)`

Answer

B

Explanation

`J=(ML^(2))/(12) and l=MK_(1)^(2)` (where `K_(1)` radius of gravitation)
`therefore" "MK_(1)^(2) rArr K_(1)=(L)/(2sqrt3)" …(i)"`
When axis rotation of rod is passing through one and of rod, then
`l=MK_(2)^(2)=(ML^(2))/(3) rArr K_(2)=(L)/(sqrt3)" ...(ii)"`
`therefore" "(K_(1))/(K_(2))=(L//2sqrt3)/(L//sqrt3)=(1)/(2)`