The moment of inertia of a solid sphere about an axis passing through the centre radius is `2/5MR^(2)` , then its radius of gyration about a parallel axis t a distance `2R` from first axis is

Options

A. `5R`
B. `sqrt(22/5)R`
C. `5/2R`
D. `sqrt(12/5)R`

Answer

b

Explanation

According to the theroem of parallel axis,
`I=I_(CG)+M(2R)^(2)`
where `I=I_(CG)=MI` about an axis through the centre of gravity.
`I=2/5 MR^(2) +4MR^(2)=22/5 MR^(2)`
or `MK^(2)=22/5MR^(2)impliesK=sqrt(22/5)R`