A point mass `m_(A)` is connected to a point mass mg by a massless rod of length / as shown in the figure. It is observed that the ratio of the moment of inertia of the system about the two axes BB and AA, which are parallel to each other and perpendicular to the rod is `I_(B B)/(I_(A A)) = 3 `
The distance of the centre of mass of the system from the mass A is

Options

A. `(1)/(2) `l
B. `(1)/(4)` l
C. `(2)/(3)` l
D. `(3)/(4)` l

Answer

B

Explanation

`I_(A A) = m_(B) l^(2)` and `I_( B B) = m_(A) l^(2)`
`(I_(B B))/(I_(A A)) = 3` (Given) `therefore (m_(A))/(m_(B)) = 3 " " … (i)`
Let x be the distance of centre of mass from mass A .
`therefore m_(A) x = m_(B) ( l - x) or (m_(A))/(m_(B)) x = l - x`
or `3x = l - x ` or `x = (l)/(4) " "` (Using (i))