The moment of inertia of a solid sphere about an axis passing through the centre radius is `1/2MR^(2)` , then its radius of gyration about a parallel axis at a distance `2R` from first axis is

Options

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

Answer

B

Explanation

As `l=MK^(2)=SigmaMR^(2)`
where, M is the total mass of the body.
This means that `K=sqrt(((l)/(M)))`

According the theorem of parallel axes, `l=l_("CG")+M(2R)^(2)` where, `l_("CG")` is moment of inertia about an axis through centre of gravity.
`therefore" "l=(2)/(5)MR^(2)+4MR^(2)`
`=(22)/(5)MR^(2)`
`rArr" "MK^(2)=(22)/(5)MR^(2)`
`therefore" "K=sqrt((22)/(5))R`