The moment of inertia of a sphere of mass M and radius R about an axis passing through its centre is `(2)/(5) MR^(2)` . The radius of gyration of the sphere about a parallel axis to the above and tangent to the sphere is

Options

A. `(7)/(5)` R
B. `(3)/(5)` R
C. `(sqrt((7)/(5)))R`
D. `(sqrt((3)/(5)))R`

Answer

C

Explanation

MOI about tangent
`I=I_(CM)+MR^(2)=(2)/(5)MR^(2)+MR^(2)=(7)/(5)MR^(2)`
Now, `(7)/(5)MR^(2)=MK^(2) impliesK=sqrt((7)/(5))R`